Related
I want to use an optimizer within the forward pass of a custom defined Function, but it doesn't work. My code is as follows:
class MyFct(Function):
#staticmethod
def forward(ctx, *args):
input, weight, bias = args[0], args[1], args[2]
y = torch.tensor([[0]], dtype=torch.float, requires_grad=True) #initial guess
loss_fn = lambda y_star: (input + weight - y_star)**2
learning_rate = 1e-4
optimizer = torch.optim.Adam([y], lr=learning_rate)
for t in range(5000):
y_star = y
print(y_star)
loss = loss_fn(y_star)
if t % 100 == 99:
print(t, loss.item())
optimizer.zero_grad()
loss.backward()
optimizer.step()
return y_star
And that's my test inputs:
x = torch.tensor([[2]], dtype=torch.float, requires_grad=True)
w = torch.tensor([[2]], dtype=torch.float, requires_grad=True)
y = torch.tensor([[6]], dtype=torch.float)
fct= MyFct.apply
y_hat = fct(x, w, None)
I always get the RuntimeError: element 0 of tensors does not require grad and does not have a grad_fn.
Also, I've tested the optimization outside of the forward and it works, so I guess it's something with the context? According to the documentation "Tensor arguments that track history (i.e., with requires_grad=True) will be converted to ones that don’t track history before the call, and their use will be registered in the graph", see https://pytorch.org/docs/stable/notes/extending.html. Is this the problem? Is there a way to work around it?
I am new to PyTorch and I wonder what I'm overlooking. Any help and explanation is appreciated.
I think I found an answer here: https://github.com/pytorch/pytorch/issues/8847 , i.e. I need to wrap the oprimization with with torch.enable_grad():.
However, I still don't understand why it's necessary to convert the original Tensors to ones that don’t track history in forward().
I am trying to solve the 1D time dependent Schroedinger equation using finite difference methods, here is how the equation looks and how it undergoes discretization
Say I have N spatial points (the x_i goes from 0 to N-1), and suppose my time span is K time points.
I strive to get a K by N matrix. each row (j) will be the function at time t_j
I suspect that my issue is that I am defining the system of the coupled equations in a wrong way.
My boundary conditions are psi=0, or some constant at the sides of the box so I make the ode's in the sides of my X span to be zero
My Code:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp
#Defining the length and the resolution of our x vector
length = 2*np.pi
delta_x = .01
# create a vector of X values, and the number of X values
def create_x_vector(length, delta_x):
x = np.arange(-length, length, delta_x)
N = len(x)
return x, N
# create initial condition vector
def create_initial_cond(x,x0,Gausswidth):
psi0 = np.exp((-(x-x0)**2)/Gausswidth)
return psi0
#create the system of ODEs
def ode_system(psi,t,delta_x,N):
psi_t = np.zeros(N)
psi_t[0]=0
psi_t[N-1]=0
for i in range(1,N-1):
psi_t[i] = (psi[i+1]-2*psi[i]+psi[i-1])/(delta_x)**2
return psi_t
#Create the actual time, x and initial condition vectors using the functions
t = np.linspace(0,15,5000)
x,N= create_x_vector(length,delta_x)
psi0 = create_initial_cond(x,0,1)
psi = np.zeros(N)
psi= solve_ivp(ode_system(psi,t,delta_x,N),[0,15],psi0,method='Radau',max_step=0.1)
After running I get an error:
runfile('D:/Studies/Project/Simulation Test/Test2.py', wdir='D:/Studies/Project/Simulation Test')
Traceback (most recent call last):
File "<ipython-input-16-bff0a1fd9937>", line 1, in <module>
runfile('D:/Studies/Project/Simulation Test/Test2.py', wdir='D:/Studies/Project/Simulation Test')
File "C:\Users\Pasha\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 704, in runfile
execfile(filename, namespace)
File "C:\Users\Pasha\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 108, in execfile
exec(compile(f.read(), filename, 'exec'), namespace)
File "D:/Studies/Project/Simulation Test/Test2.py", line 35, in <module>
psi= solve_ivp(ode_system(psi,t,delta_x,N),[0,15],psi0,method='Radau',max_step=0.1)
File "C:\Users\Pasha\Anaconda3\lib\site-packages\scipy\integrate\_ivp\ivp.py", line 454, in solve_ivp
solver = method(fun, t0, y0, tf, vectorized=vectorized, **options)
File "C:\Users\Pasha\Anaconda3\lib\site-packages\scipy\integrate\_ivp\radau.py", line 288, in __init__
self.f = self.fun(self.t, self.y)
File "C:\Users\Pasha\Anaconda3\lib\site-packages\scipy\integrate\_ivp\base.py", line 139, in fun
return self.fun_single(t, y)
File "C:\Users\Pasha\Anaconda3\lib\site-packages\scipy\integrate\_ivp\base.py", line 21, in fun_wrapped
return np.asarray(fun(t, y), dtype=dtype)
TypeError: 'numpy.ndarray' object is not callable
In a more general note, how can I make python solve N ode's without manually defining each and one of them?
I want to have a big vector called xdot where each cell in this vector will be a function of some X[i]'s and I seem to fail to do that? or maybe my approach is completely wrong?
Also I feel maybe that the "Vectorized" argument of ivp_solve can be connected, but I do not understand the explanation in the SciPy documentation.
The problem is probably that solve_ivp expects a function for its first parameter, and you provided ode_system(psi,t,delta_x,N) which results in a matrix instead (therefore you get type error - ndarray).
You need to provide solve_ivp a function that accepts two variables, t and y (which in your case is psi). It can be done like this:
def temp_function(t, psi):
return ode_system(psi,t,delta_x,N)
and then, your last line should be:
psi= solve_ivp(temp_function,[0,15],psi0,method='Radau',max_step=0.1)
This code solved the problem for me.
For a shorthand way of doing this, you can also just write the function inline using Lambda :
psi= solve_ivp(lambda t,psi : ode_system(psi,t,delta_x,N),[0,15],psi0,method='Radau',max_step=0.1)
What does the # symbol do in Python?
An # symbol at the beginning of a line is used for class and function decorators:
PEP 318: Decorators
Python Decorators
The most common Python decorators are:
#property
#classmethod
#staticmethod
An # in the middle of a line is probably matrix multiplication:
# as a binary operator.
Example
class Pizza(object):
def __init__(self):
self.toppings = []
def __call__(self, topping):
# When using '#instance_of_pizza' before a function definition
# the function gets passed onto 'topping'.
self.toppings.append(topping())
def __repr__(self):
return str(self.toppings)
pizza = Pizza()
#pizza
def cheese():
return 'cheese'
#pizza
def sauce():
return 'sauce'
print pizza
# ['cheese', 'sauce']
This shows that the function/method/class you're defining after a decorator is just basically passed on as an argument to the function/method immediately after the # sign.
First sighting
The microframework Flask introduces decorators from the very beginning in the following format:
from flask import Flask
app = Flask(__name__)
#app.route("/")
def hello():
return "Hello World!"
This in turn translates to:
rule = "/"
view_func = hello
# They go as arguments here in 'flask/app.py'
def add_url_rule(self, rule, endpoint=None, view_func=None, **options):
pass
Realizing this finally allowed me to feel at peace with Flask.
In Python 3.5 you can overload # as an operator. It is named as __matmul__, because it is designed to do matrix multiplication, but it can be anything you want. See PEP465 for details.
This is a simple implementation of matrix multiplication.
class Mat(list):
def __matmul__(self, B):
A = self
return Mat([[sum(A[i][k]*B[k][j] for k in range(len(B)))
for j in range(len(B[0])) ] for i in range(len(A))])
A = Mat([[1,3],[7,5]])
B = Mat([[6,8],[4,2]])
print(A # B)
This code yields:
[[18, 14], [62, 66]]
This code snippet:
def decorator(func):
return func
#decorator
def some_func():
pass
Is equivalent to this code:
def decorator(func):
return func
def some_func():
pass
some_func = decorator(some_func)
In the definition of a decorator you can add some modified things that wouldn't be returned by a function normally.
What does the “at” (#) symbol do in Python?
In short, it is used in decorator syntax and for matrix multiplication.
In the context of decorators, this syntax:
#decorator
def decorated_function():
"""this function is decorated"""
is equivalent to this:
def decorated_function():
"""this function is decorated"""
decorated_function = decorator(decorated_function)
In the context of matrix multiplication, a # b invokes a.__matmul__(b) - making this syntax:
a # b
equivalent to
dot(a, b)
and
a #= b
equivalent to
a = dot(a, b)
where dot is, for example, the numpy matrix multiplication function and a and b are matrices.
How could you discover this on your own?
I also do not know what to search for as searching Python docs or Google does not return relevant results when the # symbol is included.
If you want to have a rather complete view of what a particular piece of python syntax does, look directly at the grammar file. For the Python 3 branch:
~$ grep -C 1 "#" cpython/Grammar/Grammar
decorator: '#' dotted_name [ '(' [arglist] ')' ] NEWLINE
decorators: decorator+
--
testlist_star_expr: (test|star_expr) (',' (test|star_expr))* [',']
augassign: ('+=' | '-=' | '*=' | '#=' | '/=' | '%=' | '&=' | '|=' | '^=' |
'<<=' | '>>=' | '**=' | '//=')
--
arith_expr: term (('+'|'-') term)*
term: factor (('*'|'#'|'/'|'%'|'//') factor)*
factor: ('+'|'-'|'~') factor | power
We can see here that # is used in three contexts:
decorators
an operator between factors
an augmented assignment operator
Decorator Syntax:
A google search for "decorator python docs" gives as one of the top results, the "Compound Statements" section of the "Python Language Reference." Scrolling down to the section on function definitions, which we can find by searching for the word, "decorator", we see that... there's a lot to read. But the word, "decorator" is a link to the glossary, which tells us:
decorator
A function returning another function, usually applied as a function transformation using the #wrapper syntax. Common
examples for decorators are classmethod() and staticmethod().
The decorator syntax is merely syntactic sugar, the following two
function definitions are semantically equivalent:
def f(...):
...
f = staticmethod(f)
#staticmethod
def f(...):
...
The same concept exists for classes, but is less commonly used there.
See the documentation for function definitions and class definitions
for more about decorators.
So, we see that
#foo
def bar():
pass
is semantically the same as:
def bar():
pass
bar = foo(bar)
They are not exactly the same because Python evaluates the foo expression (which could be a dotted lookup and a function call) before bar with the decorator (#) syntax, but evaluates the foo expression after bar in the other case.
(If this difference makes a difference in the meaning of your code, you should reconsider what you're doing with your life, because that would be pathological.)
Stacked Decorators
If we go back to the function definition syntax documentation, we see:
#f1(arg)
#f2
def func(): pass
is roughly equivalent to
def func(): pass
func = f1(arg)(f2(func))
This is a demonstration that we can call a function that's a decorator first, as well as stack decorators. Functions, in Python, are first class objects - which means you can pass a function as an argument to another function, and return functions. Decorators do both of these things.
If we stack decorators, the function, as defined, gets passed first to the decorator immediately above it, then the next, and so on.
That about sums up the usage for # in the context of decorators.
The Operator, #
In the lexical analysis section of the language reference, we have a section on operators, which includes #, which makes it also an operator:
The following tokens are operators:
+ - * ** / // % #
<< >> & | ^ ~
< > <= >= == !=
and in the next page, the Data Model, we have the section Emulating Numeric Types,
object.__add__(self, other)
object.__sub__(self, other)
object.__mul__(self, other)
object.__matmul__(self, other)
object.__truediv__(self, other)
object.__floordiv__(self, other)
[...]
These methods are called to implement the binary arithmetic operations (+, -, *, #, /, //, [...]
And we see that __matmul__ corresponds to #. If we search the documentation for "matmul" we get a link to What's new in Python 3.5 with "matmul" under a heading "PEP 465 - A dedicated infix operator for matrix multiplication".
it can be implemented by defining __matmul__(), __rmatmul__(), and
__imatmul__() for regular, reflected, and in-place matrix multiplication.
(So now we learn that #= is the in-place version). It further explains:
Matrix multiplication is a notably common operation in many fields of
mathematics, science, engineering, and the addition of # allows
writing cleaner code:
S = (H # beta - r).T # inv(H # V # H.T) # (H # beta - r)
instead of:
S = dot((dot(H, beta) - r).T,
dot(inv(dot(dot(H, V), H.T)), dot(H, beta) - r))
While this operator can be overloaded to do almost anything, in numpy, for example, we would use this syntax to calculate the inner and outer product of arrays and matrices:
>>> from numpy import array, matrix
>>> array([[1,2,3]]).T # array([[1,2,3]])
array([[1, 2, 3],
[2, 4, 6],
[3, 6, 9]])
>>> array([[1,2,3]]) # array([[1,2,3]]).T
array([[14]])
>>> matrix([1,2,3]).T # matrix([1,2,3])
matrix([[1, 2, 3],
[2, 4, 6],
[3, 6, 9]])
>>> matrix([1,2,3]) # matrix([1,2,3]).T
matrix([[14]])
Inplace matrix multiplication: #=
While researching the prior usage, we learn that there is also the inplace matrix multiplication. If we attempt to use it, we may find it is not yet implemented for numpy:
>>> m = matrix([1,2,3])
>>> m #= m.T
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: In-place matrix multiplication is not (yet) supported. Use 'a = a # b' instead of 'a #= b'.
When it is implemented, I would expect the result to look like this:
>>> m = matrix([1,2,3])
>>> m #= m.T
>>> m
matrix([[14]])
What does the “at” (#) symbol do in Python?
# symbol is a syntactic sugar python provides to utilize decorator,
to paraphrase the question, It's exactly about what does decorator do in Python?
Put it simple decorator allow you to modify a given function's definition without touch its innermost (it's closure).
It's the most case when you import wonderful package from third party. You can visualize it, you can use it, but you cannot touch its innermost and its heart.
Here is a quick example,
suppose I define a read_a_book function on Ipython
In [9]: def read_a_book():
...: return "I am reading the book: "
...:
In [10]: read_a_book()
Out[10]: 'I am reading the book: '
You see, I forgot to add a name to it.
How to solve such a problem? Of course, I could re-define the function as:
def read_a_book():
return "I am reading the book: 'Python Cookbook'"
Nevertheless, what if I'm not allowed to manipulate the original function, or if there are thousands of such function to be handled.
Solve the problem by thinking different and define a new_function
def add_a_book(func):
def wrapper():
return func() + "Python Cookbook"
return wrapper
Then employ it.
In [14]: read_a_book = add_a_book(read_a_book)
In [15]: read_a_book()
Out[15]: 'I am reading the book: Python Cookbook'
Tada, you see, I amended read_a_book without touching it inner closure. Nothing stops me equipped with decorator.
What's about #
#add_a_book
def read_a_book():
return "I am reading the book: "
In [17]: read_a_book()
Out[17]: 'I am reading the book: Python Cookbook'
#add_a_book is a fancy and handy way to say read_a_book = add_a_book(read_a_book), it's a syntactic sugar, there's nothing more fancier about it.
If you are referring to some code in a python notebook which is using Numpy library, then # operator means Matrix Multiplication. For example:
import numpy as np
def forward(xi, W1, b1, W2, b2):
z1 = W1 # xi + b1
a1 = sigma(z1)
z2 = W2 # a1 + b2
return z2, a1
Decorators were added in Python to make function and method wrapping (a function that receives a function and returns an enhanced one) easier to read and understand. The original use case was to be able to define the methods as class methods or static methods on the head of their definition. Without the decorator syntax, it would require a rather sparse and repetitive definition:
class WithoutDecorators:
def some_static_method():
print("this is static method")
some_static_method = staticmethod(some_static_method)
def some_class_method(cls):
print("this is class method")
some_class_method = classmethod(some_class_method)
If the decorator syntax is used for the same purpose, the code is shorter and easier to understand:
class WithDecorators:
#staticmethod
def some_static_method():
print("this is static method")
#classmethod
def some_class_method(cls):
print("this is class method")
General syntax and possible implementations
The decorator is generally a named object ( lambda expressions are not allowed) that accepts a single argument when called (it will be the decorated function) and returns another callable object. "Callable" is used here instead of "function" with premeditation. While decorators are often discussed in the scope of methods and functions, they are not limited to them. In fact, anything that is callable (any object that implements the _call__ method is considered callable), can be used as a decorator and often objects returned by them are not simple functions but more instances of more complex classes implementing their own __call_ method.
The decorator syntax is simply only a syntactic sugar. Consider the following decorator usage:
#some_decorator
def decorated_function():
pass
This can always be replaced by an explicit decorator call and function reassignment:
def decorated_function():
pass
decorated_function = some_decorator(decorated_function)
However, the latter is less readable and also very hard to understand if multiple decorators are used on a single function.
Decorators can be used in multiple different ways as shown below:
As a function
There are many ways to write custom decorators, but the simplest way is to write a function that returns a subfunction that wraps the original function call.
The generic patterns is as follows:
def mydecorator(function):
def wrapped(*args, **kwargs):
# do some stuff before the original
# function gets called
result = function(*args, **kwargs)
# do some stuff after function call and
# return the result
return result
# return wrapper as a decorated function
return wrapped
As a class
While decorators almost always can be implemented using functions, there are some situations when using user-defined classes is a better option. This is often true when the decorator needs complex parametrization or it depends on a specific state.
The generic pattern for a nonparametrized decorator as a class is as follows:
class DecoratorAsClass:
def __init__(self, function):
self.function = function
def __call__(self, *args, **kwargs):
# do some stuff before the original
# function gets called
result = self.function(*args, **kwargs)
# do some stuff after function call and
# return the result
return result
Parametrizing decorators
In real code, there is often a need to use decorators that can be parametrized. When the function is used as a decorator, then the solution is simple—a second level of wrapping has to be used. Here is a simple example of the decorator that repeats the execution of a decorated function the specified number of times every time it is called:
def repeat(number=3):
"""Cause decorated function to be repeated a number of times.
Last value of original function call is returned as a result
:param number: number of repetitions, 3 if not specified
"""
def actual_decorator(function):
def wrapper(*args, **kwargs):
result = None
for _ in range(number):
result = function(*args, **kwargs)
return result
return wrapper
return actual_decorator
The decorator defined this way can accept parameters:
>>> #repeat(2)
... def foo():
... print("foo")
...
>>> foo()
foo
foo
Note that even if the parametrized decorator has default values for its arguments, the parentheses after its name is required. The correct way to use the preceding decorator with default arguments is as follows:
>>> #repeat()
... def bar():
... print("bar")
...
>>> bar()
bar
bar
bar
Finally lets see decorators with Properties.
Properties
The properties provide a built-in descriptor type that knows how to link an attribute to a set of methods. A property takes four optional arguments: fget , fset , fdel , and doc . The last one can be provided to define a docstring that is linked to the attribute as if it were a method. Here is an example of a Rectangle class that can be controlled either by direct access to attributes that store two corner points or by using the width , and height properties:
class Rectangle:
def __init__(self, x1, y1, x2, y2):
self.x1, self.y1 = x1, y1
self.x2, self.y2 = x2, y2
def _width_get(self):
return self.x2 - self.x1
def _width_set(self, value):
self.x2 = self.x1 + value
def _height_get(self):
return self.y2 - self.y1
def _height_set(self, value):
self.y2 = self.y1 + value
width = property(
_width_get, _width_set,
doc="rectangle width measured from left"
)
height = property(
_height_get, _height_set,
doc="rectangle height measured from top"
)
def __repr__(self):
return "{}({}, {}, {}, {})".format(
self.__class__.__name__,
self.x1, self.y1, self.x2, self.y2
)
The best syntax for creating properties is using property as a decorator. This will reduce the number of method signatures inside of the class
and make code more readable and maintainable. With decorators the above class becomes:
class Rectangle:
def __init__(self, x1, y1, x2, y2):
self.x1, self.y1 = x1, y1
self.x2, self.y2 = x2, y2
#property
def width(self):
"""rectangle height measured from top"""
return self.x2 - self.x1
#width.setter
def width(self, value):
self.x2 = self.x1 + value
#property
def height(self):
"""rectangle height measured from top"""
return self.y2 - self.y1
#height.setter
def height(self, value):
self.y2 = self.y1 + value
Starting with Python 3.5, the '#' is used as a dedicated infix symbol for MATRIX MULTIPLICATION (PEP 0465 -- see https://www.python.org/dev/peps/pep-0465/)
# can be a math operator or a DECORATOR but what you mean is a decorator.
This code:
def func(f):
return f
func(lambda :"HelloWorld")()
using decorators can be written like:
def func(f):
return f
#func
def name():
return "Hello World"
name()
Decorators can have arguments.
You can see this GeeksforGeeks post: https://www.geeksforgeeks.org/decorators-in-python/
It indicates that you are using a decorator. Here is Bruce Eckel's example from 2008.
Python decorator is like a wrapper of a function or a class. It’s still too conceptual.
def function_decorator(func):
def wrapped_func():
# Do something before the function is executed
func()
# Do something after the function has been executed
return wrapped_func
The above code is a definition of a decorator that decorates a function.
function_decorator is the name of the decorator.
wrapped_func is the name of the inner function, which is actually only used in this decorator definition. func is the function that is being decorated.
In the inner function wrapped_func, we can do whatever before and after the func is called. After the decorator is defined, we simply use it as follows.
#function_decorator
def func():
pass
Then, whenever we call the function func, the behaviours we’ve defined in the decorator will also be executed.
EXAMPLE :
from functools import wraps
def mydecorator(f):
#wraps(f)
def wrapped(*args, **kwargs):
print "Before decorated function"
r = f(*args, **kwargs)
print "After decorated function"
return r
return wrapped
#mydecorator
def myfunc(myarg):
print "my function", myarg
return "return value"
r = myfunc('asdf')
print r
Output :
Before decorated function
my function asdf
After decorated function
return value
To say what others have in a different way: yes, it is a decorator.
In Python, it's like:
Creating a function (follows under the # call)
Calling another function to operate on your created function. This returns a new function. The function that you call is the argument of the #.
Replacing the function defined with the new function returned.
This can be used for all kinds of useful things, made possible because functions are objects and just necessary just instructions.
# symbol is also used to access variables inside a plydata / pandas dataframe query, pandas.DataFrame.query.
Example:
df = pandas.DataFrame({'foo': [1,2,15,17]})
y = 10
df >> query('foo > #y') # plydata
df.query('foo > #y') # pandas
I have a function f that is internally using some tf.while_loops and tf.gradients to compute the value y = f(x). Something like this
def f( x ):
...
def body( g, x ):
# Compute the gradient here
grad = tf.gradients( g, x )[0]
...
return ...
return tf.while_loop( cond, body, parallel_iterations=1 )
There are a few hundred lines of code. But I believe that those are the important points...
Now when I evaluate f(x), I get exactly the value I expect ..
y = known output of f(x)
with tf.Session() as sess:
fx = f(x)
print("Error = ", y - sess.run(fx, feed_dict)) # Prints 0
However, when I try to evaluate the gradient of f(x) with respect to x, that is,
grads = tf.gradients( fx, x )[0]
I get the error
AssertionError: gradients list should have been aggregated by now.
Here is the full trace:
File "C:/Dropbox/bob/tester.py", line 174, in <module>
grads = tf.gradients(y, x)[0]
File "C:\Anaconda36\lib\site-packages\tensorflow\python\ops\gradients_impl.py", line 649, in gradients
return [_GetGrad(grads, x) for x in xs]
File "C:\Anaconda36\lib\site-packages\tensorflow\python\ops\gradients_impl.py", line 649, in <listcomp>
return [_GetGrad(grads, x) for x in xs]
File "C:\Anaconda36\lib\site-packages\tensorflow\python\ops\gradients_impl.py", line 727, in _GetGrad
"gradients list should have been aggregated by now.")
AssertionError: gradients list should have been aggregated by now.
Could somebody please outline likely causes for this error? I have no idea where to even start looking for the issue...
Some observations:
Note that I have set the parallel iterations for the while loop to 1. This
should mean that there is no errors due to reading and writing from multiple threads.
If I discard the while loop, and just have f return body(), then the code runs:
# The following does not crash, but we removed the while_loop, so the output is incorrect
def f( x ):
...
def body( g, x ):
# Compute the gradient here
grad = tf.gradients( g, x )[0]
...
return ...
return body(...)
Obviously, the output is incorrect, but at least the gradients are computed.
I came across a similar issue. Some patterns I noted:
If the x used in tf.gradients was used in a manner that required dimension broadcasting in body, I got this error. If I changed it to one that didn't require broadcasting, tf.gradients returned [None]. I didn't test this extensively, so this pattern may not be consistent across all examples.
Both cases (returning [None] and raising this assertion error) can be resolved by differentiating tf.identity(y) rather than just y: grads = tf.gradients(tf.identity(y), xs) I have absolutely no idea why this works.
Strange error from numpy via matplotlib when trying to get a histogram of a tiny toy dataset. I'm just not sure how to interpret the error, which makes it hard to see what to do next.
Didn't find much related, though this nltk question and this gdsCAD question are superficially similar.
I intend the debugging info at bottom to be more helpful than the driver code, but if I've missed something, please ask. This is reproducible as part of an existing test suite.
if n > 1:
return diff(a[slice1]-a[slice2], n-1, axis=axis)
else:
> return a[slice1]-a[slice2]
E TypeError: ufunc 'subtract' did not contain a loop with signature matching types dtype('<U1') dtype('<U1') dtype('<U1')
../py2.7.11-venv/lib/python2.7/site-packages/numpy/lib/function_base.py:1567: TypeError
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> entering PDB >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
> py2.7.11-venv/lib/python2.7/site-packages/numpy/lib/function_base.py(1567)diff()
-> return a[slice1]-a[slice2]
(Pdb) bt
[...]
py2.7.11-venv/lib/python2.7/site-packages/matplotlib/axes/_axes.py(5678)hist()
-> m, bins = np.histogram(x[i], bins, weights=w[i], **hist_kwargs)
py2.7.11-venv/lib/python2.7/site-packages/numpy/lib/function_base.py(606)histogram()
-> if (np.diff(bins) < 0).any():
> py2.7.11-venv/lib/python2.7/site-packages/numpy/lib/function_base.py(1567)diff()
-> return a[slice1]-a[slice2]
(Pdb) p numpy.__version__
'1.11.0'
(Pdb) p matplotlib.__version__
'1.4.3'
(Pdb) a
a = [u'A' u'B' u'C' u'D' u'E']
n = 1
axis = -1
(Pdb) p slice1
(slice(1, None, None),)
(Pdb) p slice2
(slice(None, -1, None),)
(Pdb)
I got the same error, but in my case I am subtracting dict.key from dict.value. I have fixed this by subtracting dict.value for corresponding key from other dict.value.
cosine_sim = cosine_similarity(e_b-e_a, w-e_c)
here I got error because e_b, e_a and e_c are embedding vector for word a,b,c respectively. I didn't know that 'w' is string, when I sought out w is string then I fix this by following line:
cosine_sim = cosine_similarity(e_b-e_a, word_to_vec_map[w]-e_c)
Instead of subtracting dict.key, now I have subtracted corresponding value for key
I had a similar issue where an integer in a row of a DataFrame I was iterating over was of type numpy.int64. I got the
TypeError: ufunc 'subtract' did not contain a loop with signature matching types dtype('<U1') dtype('<U1') dtype('<U1')
error when trying to subtract a float from it.
The easiest fix for me was to convert the row using pd.to_numeric(row).
Why is it applying diff to an array of strings.
I get an error at the same point, though with a different message
In [23]: a=np.array([u'A' u'B' u'C' u'D' u'E'])
In [24]: np.diff(a)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-24-9d5a62fc3ff0> in <module>()
----> 1 np.diff(a)
C:\Users\paul\AppData\Local\Enthought\Canopy\User\lib\site-packages\numpy\lib\function_base.pyc in diff(a, n, axis)
1112 return diff(a[slice1]-a[slice2], n-1, axis=axis)
1113 else:
-> 1114 return a[slice1]-a[slice2]
1115
1116
TypeError: unsupported operand type(s) for -: 'numpy.ndarray' and 'numpy.ndarray'
Is this a array the bins parameter? What does the docs say bins should be?
I am fairly new to this myself, but I had a similar error and found that it is due to a type casting issue. I was trying to concatenate rather than take the difference but I think the principle is the same here. I provided a similar answer on another question so I hope that is OK.
In essence you need to use a different data type cast, in my case I needed str not float, I suspect yours is the same so my suggested solution is. I am sorry I cannot test it before suggesting but I am unclear from your example what you were doing.
return diff(str(a[slice1])-str(a[slice2]), n-1, axis=axis)
Please see my example code below for the fix to my code, the change occurs on the third to last line. The code is to produce a basic random forest model.
import scipy
import math
import numpy as np
import pandas as pd
from sklearn.ensemble import RandomForestRegressor
from sklearn import preprocessing, metrics, cross_validation
Data = pd.read_csv("Free_Energy_exp.csv", sep=",")
Data = Data.fillna(Data.mean()) # replace the NA values with the mean of the descriptor
header = Data.columns.values # Ues the column headers as the descriptor labels
Data.head()
test_name = "Test.csv"
npArray = np.array(Data)
print header.shape
npheader = np.array(header[1:-1])
print("Array shape X = %d, Y = %d " % (npArray.shape))
datax, datay = npArray.shape
names = npArray[:,0]
X = npArray[:,1:-1].astype(float)
y = npArray[:,-1] .astype(float)
X = preprocessing.scale(X)
XTrain, XTest, yTrain, yTest = cross_validation.train_test_split(X,y, random_state=0)
# Predictions results initialised
RFpredictions = []
RF = RandomForestRegressor(n_estimators = 10, max_features = 5, max_depth = 5, random_state=0)
RF.fit(XTrain, yTrain) # Train the model
print("Training R2 = %5.2f" % RF.score(XTrain,yTrain))
RFpreds = RF.predict(XTest)
with open(test_name,'a') as fpred :
lenpredictions = len(RFpreds)
lentrue = yTest.shape[0]
if lenpredictions == lentrue :
fpred.write("Names/Label,, Prediction Random Forest,, True Value,\n")
for i in range(0,lenpredictions) :
fpred.write(RFpreds[i]+",,"+yTest[i]+",\n")
else :
print "ERROR - names, prediction and true value array size mismatch."
This leads to an error of;
Traceback (most recent call last):
File "min_example.py", line 40, in <module>
fpred.write(RFpreds[i]+",,"+yTest[i]+",\n")
TypeError: ufunc 'add' did not contain a loop with signature matching types dtype('S32') dtype('S32') dtype('S32')
The solution is to make each variable a str() type on the third to last line then write to file. No other changes to then code have been made from the above.
import scipy
import math
import numpy as np
import pandas as pd
from sklearn.ensemble import RandomForestRegressor
from sklearn import preprocessing, metrics, cross_validation
Data = pd.read_csv("Free_Energy_exp.csv", sep=",")
Data = Data.fillna(Data.mean()) # replace the NA values with the mean of the descriptor
header = Data.columns.values # Ues the column headers as the descriptor labels
Data.head()
test_name = "Test.csv"
npArray = np.array(Data)
print header.shape
npheader = np.array(header[1:-1])
print("Array shape X = %d, Y = %d " % (npArray.shape))
datax, datay = npArray.shape
names = npArray[:,0]
X = npArray[:,1:-1].astype(float)
y = npArray[:,-1] .astype(float)
X = preprocessing.scale(X)
XTrain, XTest, yTrain, yTest = cross_validation.train_test_split(X,y, random_state=0)
# Predictions results initialised
RFpredictions = []
RF = RandomForestRegressor(n_estimators = 10, max_features = 5, max_depth = 5, random_state=0)
RF.fit(XTrain, yTrain) # Train the model
print("Training R2 = %5.2f" % RF.score(XTrain,yTrain))
RFpreds = RF.predict(XTest)
with open(test_name,'a') as fpred :
lenpredictions = len(RFpreds)
lentrue = yTest.shape[0]
if lenpredictions == lentrue :
fpred.write("Names/Label,, Prediction Random Forest,, True Value,\n")
for i in range(0,lenpredictions) :
fpred.write(str(RFpreds[i])+",,"+str(yTest[i])+",\n")
else :
print "ERROR - names, prediction and true value array size mismatch."
These examples are from a larger code so I hope the examples are clear enough.
I think #James is right. I got stuck by same error while working on Polyval(). And yeah solution is to use the same type of variabes. You can use typecast to cast all variables in the same type.
BELOW IS A EXAMPLE CODE
import numpy
P = numpy.array(input().split(), float)
x = float(input())
print(numpy.polyval(P,x))
here I used float as an output type. so even the user inputs the INT value (whole number). the final answer will be typecasted to float.
I ran into the same issue, but in my case it was just a Python list instead of a Numpy array used. Using two Numpy arrays solved the issue for me.