I'm trying to create a TensorFlow (2.0) variable like this:
c_init = tf.zeros_initializer()
c = tf.Variable(initial_value=c_init(shape=shape, dtype="float32"), trainable=True)
the shape variable is this:
shape=(49, 52, 26, 49, 6, 3, 31, 11, 24, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)
I'm getting this error message:
InvalidArgumentError: Too many dimensions [Op:Fill] name: zeros/
I did not know there is a limit in the number of dimensions. I did not see anything in TensorFlow documentation about it. Is there any way to get around this limitation?
The max number of dimensions is 254
Get around this limitation?
If you don't mind, I do have to ask you a question:
Are you sure you have that many dimensions in your problem? I have seen people mistakenly using size of dimension as number of dimensions. Are you sure you have so many dimensions with size 1?
Let's not forget that there is no need to represent anything as tensors. We could solve any problem without using a multi-dimensional data type (tensor). The reason why this type of representation is used is because it allows certain linear algebra operations to be applied and they are much fast when compared to regular loops in a more traditional code.
So, yes, you can get around this limitation but you will need to so some "soul search" and figure out what kind of math operations you are planning to apply to this humongous tensor.
Related
Edit: example DataFrame for the original error-message found and posted.
(As I just recognized, the Error does only appear, if the tuple has a certain length. The example is now adapted.)
Original text:
I need to group by tuple of different length.
For the grouping I'm applying a summary_function.
import pandas as pd
def summary_function(df):
value_mean = df['value'].mean()
df1 = pd.DataFrame({'value_mean':[value_mean]
})
return df1
tuple_list = [(1,2,1,1,1,1,1,1,1,1,1,1,1),(2,3,1,1,1,1,1,1,1,1,1,1,1), \
(1,2,1,1,1,1,1,1,1,1,1,1,1), \
(2,3,4,4,4,4,4,4,4,4,4,4,4,4,4,1,1,1,1,1,1,1,1,1,1,1)]
value = [1,2,3,4]
letter = list('abab')
df = pd.DataFrame({'letter':letter, 'tuple':tuple_list, 'value':value})
df
> letter tuple value
>0 a (1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 1
>1 b (2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 2
>2 a (1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 3
>3 b (2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ... 4
If I'm using a direct mean() function, the result is how expected:
df.groupby(['letter','tuple']).mean()
> value
>letter tuple
>a (1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 2
>b (2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 2
> (2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ...) 4
But if I apply the function. (which I need to use since I have dozens of summaries) The tupel is empty while using the simple
df.groupby(['letter','tuple']).apply(lambda x:summary_function(x))
I get a ValueError:
>ValueError: Values not found in passed level: MultiIndex([(2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4)],
)
It would be awesome to get some ideas on how to solve this.
In your case, do not return the dataframe, return the series.
When you return the series, Pandas will align the series horizontally. For example:
def summary_function(df):
return df['value'].agg(['min','mean','max'])
df.groupby(['letter','tuple']).apply(summary_function)
Output:
value min mean max
letter tuple
a (1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 1.0 2.0 3.0
b (2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) 2.0 2.0 2.0
(2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 1... 4.0 4.0 4.0
The even shorter solution was just to replace "pd.DataFrame" with "pd.Series".
def summary_function(df):
value_mean = df['value'].mean()
df1 = pd.Series({'value_mean':[value_mean]
})
(Inspired by the answer of Quang Hoang)
I want to create a matrix consists of some submatrix, or elements are defined by some conditions about indices.
e.g.
X = np.array(
[[0, 0, 1, 1, 1],
[0, 0, 1, 1, 1],
[1, 1, 0, 0, 1],
[1, 1, 0, 0, 1],
[1, 1, 1, 1, 0]]
)
where i-row and j-col meets conditions below
0 if 2k ≤ i < 2(k+1) and 2k ≤ j < 2(k+1)
1 otherwise
In the above condition, k is 0, 1, 2... and 2 is also a parameter to change and
So, what I finally need is
0 if nk ≤ i < n(k+1) nk ≤ j < n(k+1)
1 otherwise
I think np.ix_ is good for this demand, but it requires a loop structure (I hate loops).
Is there some nice way to generate this?
One way would be utilizing the outer method of the not_equal operator like so:
N = 17; k = 5
np.not_equal.outer(*2*(np.arange(N)//k,)).view('u1')
# array([[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0]], dtype=uint8)
Anyone know how to print more than 32 values? My output looks like this, and I'm trying to make it show the rest of the array:
Value of: model.GetOutput(0)
Expected: contains 64 values, where each value and its corresponding value in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, ... } are an almost-equal pair
Actual: { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... }, where the value pair (1, 2) at index #1 don't match, which is 1 from 1
It's hard-coded in the Google Test sources (kMaxCount = 32). To change it, you have to modify the code and rebuild Google Test. You might be able to define your own printer if the type is specific enough.
I am trying to compute matrix z (defined below) in python with numpy.
Here's my current solution (using 1 for loop)
z = np.zeros((n, k))
for i in range(n):
v = pi * (1 / math.factorial(x[i])) * np.exp(-1 * lamb) * (lamb ** x[i])
numerator = np.sum(v)
c = v / numerator
z[i, :] = c
return z
Is it possible to completely vectorize this computation? I need to do this computation for thousands of iterations, and matrix operations in numpy is much faster than huge for loops.
Here is a vectorized version of E. It replaces the for-loop and scalar arithmetic with NumPy broadcasting and array-based arithmetic:
def alt_E(x):
x = x[:, None]
z = pi * (np.exp(-lamb) * (lamb**x)) / special.factorial(x)
denom = z.sum(axis=1)[:, None]
z /= denom
return z
I ran em.py to get a sense for the typical size of x, lamb, pi, n and k. On data of this size,
alt_E is about 120x faster than E:
In [32]: %timeit E(x)
100 loops, best of 3: 11.5 ms per loop
In [33]: %timeit alt_E(x)
10000 loops, best of 3: 94.7 µs per loop
In [34]: 11500/94.7
Out[34]: 121.43611404435057
This is the setup I used for the benchmark:
import math
import numpy as np
import scipy.special as special
def alt_E(x):
x = x[:, None]
z = pi * (np.exp(-lamb) * (lamb**x)) / special.factorial(x)
denom = z.sum(axis=1)[:, None]
z /= denom
return z
def E(x):
z = np.zeros((n, k))
for i in range(n):
v = pi * (1 / math.factorial(x[i])) * \
np.exp(-1 * lamb) * (lamb ** x[i])
numerator = np.sum(v)
c = v / numerator
z[i, :] = c
return z
n = 576
k = 2
x = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5])
lamb = np.array([ 0.84835141, 1.04025989])
pi = np.array([ 0.5806958, 0.4193042])
assert np.allclose(alt_E(x), E(x))
By the way, E could also be calculated using scipy.stats.poisson:
import scipy.stats as stats
pois = stats.poisson(mu=lamb)
def alt_E2(x):
z = pi * pois.pmf(x[:,None])
denom = z.sum(axis=1)[:, None]
z /= denom
return z
but this does not turn out to be faster, at least for arrays of this length:
In [33]: %timeit alt_E(x)
10000 loops, best of 3: 94.7 µs per loop
In [102]: %timeit alt_E2(x)
1000 loops, best of 3: 278 µs per loop
For larger x, alt_E2 is faster:
In [104]: x = np.random.random(10000)
In [106]: %timeit alt_E(x)
100 loops, best of 3: 2.18 ms per loop
In [105]: %timeit alt_E2(x)
1000 loops, best of 3: 643 µs per loop
I'm building a graph which allows edges to be toggled on/off. I need to be able to add and remove them repeatedly. I have noticed this error with node degrees with nodes attached to toggled edges. I've included an example.
My code:
allElements = cy.elements();
....
var allEdges = allElements.filter('edge');
var allNodes = allElements.filter('node');
for(var i=0; i<5; i++){
// DELETE
var printThis = [];
allNodes.filter(function(i,ele){
printThis.push(ele.degree());
});
console.log(printThis);
cy.remove(allEdges);
cy.add(allEdges);
}
Returns:
[1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 6, 1, 2, 1, 1, 1, 36, 8, 3, 4, 4, 2, 1, 1, 1, 1, 1, 1, 2]
[1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 6, 1, 2, 1, 1, 1, 36, 8, 3, 4, 4, 2, 1, 1, 1, 1, 1, 1, 2]
[2, 2, 2, 2, 2, 6, 2, 2, 2, 2, 2, 12, 2, 4, 2, 2, 2, 72, 16, 6, 8, 8, 4, 2, 2, 2, 2, 2, 2, 4]
[3, 3, 3, 3, 3, 9, 3, 3, 3, 3, 3, 18, 3, 6, 3, 3, 3, 108, 24, 9, 12, 12, 6, 3, 3, 3, 3, 3, 3, 6]
[4, 4, 4, 4, 4, 12, 4, 4, 4, 4, 4, 24, 4, 8, 4, 4, 4, 144, 32, 12, 16, 16, 8, 4, 4, 4, 4, 4, 4, 8]
Which shows that removing edges after the first time dont decrease the degree of the nodes they're attached to.
How can I have cytoscape return the correct degree?
Thank you for notifying us of the issue. We will get a fix in for 2.0.3 -M
https://github.com/cytoscape/cytoscape.js/issues/360