Related
With the numpy array
arr = np.array([[1, 1, 0, 0, 0, 1], [1, 1, 0, 0, 1, 1], [1, 1, 0, 0, 0, 1]])
I would like to get the indices of all rows where the row slice 2:5 contains all zeros.
In the above example, it should return rows 0 and 2.
I tried:
zero_indices = np.where(not np.any(arr[:,2:5]))
but it doesn't seem to work.
I'm trying to do this over a large array with several million rows.
Try this
np.nonzero((~arr[:,2:5].astype(bool)).all(1))[0]
Out[133]: array([0, 2], dtype=int32)
Or
np.nonzero((arr[:,2:5] == 0).all(1))[0]
Out[139]: array([0, 2], dtype=int32)
I would like to add the values in the above row to the row below using vectorization. For example, if I had the ndarray,
[[0, 0, 0, 0],
[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3]]
Then after one iteration through this method, it would result in
[[0, 0, 0, 0],
[1, 1, 1, 1],
[3, 3, 3, 3],
[5, 5, 5, 5]]
One can simply do this with a for loop:
import numpy as np
def addAboveRow(arr):
cpy = arr.copy()
r, c = arr.shape
for i in range(1, r):
for j in range(c):
cpy[i][j] += arr[i - 1][j]
return cpy
ndarr = np.array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3]).reshape(4, 4)
print(addAboveRow(ndarr))
I'm not sure how to approach this using vectorization though. I think slicers should be used? Also, I'm not really sure how to deal with the issue of the top border, because nothing should be added onto the first row. Any help would be appreciated. Thanks!
Note: I am really new to vectorization so an explanation would be great!
You can use indexing directly:
b = np.zeros_like(a)
b[0] = a[0]
b[1:] = a[1:] + a[:-1]
>>> b
array([[0, 0, 0, 0],
[1, 1, 1, 1],
[3, 3, 3, 3],
[5, 5, 5, 5]])
An alternative:
b = a.copy()
b[1:] += a[:-1]
Or:
b = a.copy()
np.add(b[1:], a[:-1], out=b[1:])
You could try the following
np.put(arr, np.arange(arr.shape[1], arr.size), arr[1:]+arr[:-1])
I have my X and Y numpy arrays:
X = np.array([0,1,2,3])
Y = np.array([0,1,2,3])
And my function which maps x,y values to Z points:
def z(x,y):
return x+y
I wish to produce the obvious thing required for a 3D plot: the 2-dimensional numpy array for the corresponding Z-values. I believe it should look like:
Z = np.array([[0, 1, 2, 3],
[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6]])
I can do this in several lines, but I'm looking for the briefest most elegant piece of code.
For a function that is array aware it is more economical to use an open grid:
>>> import numpy as np
>>>
>>> X = np.array([0,1,2,3])
>>> Y = np.array([0,1,2,3])
>>>
>>> def z(x,y):
... return x+y
...
>>> XX, YY = np.ix_(X, Y)
>>> XX, YY
(array([[0],
[1],
[2],
[3]]), array([[0, 1, 2, 3]]))
>>> z(XX, YY)
array([[0, 1, 2, 3],
[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6]])
If your grid axes are ranges you can directly create the grid using np.ogrid
>>> XX, YY = np.ogrid[:4, :4]
>>> XX, YY
(array([[0],
[1],
[2],
[3]]), array([[0, 1, 2, 3]]))
If the function is not array aware you can make it so using np.vectorize:
>>> def f(x, y):
... if x > y:
... return x
... else:
... return -x
...
>>> np.vectorize(f)(*np.ogrid[-3:4, -3:4])
array([[ 3, 3, 3, 3, 3, 3, 3],
[-2, 2, 2, 2, 2, 2, 2],
[-1, -1, 1, 1, 1, 1, 1],
[ 0, 0, 0, 0, 0, 0, 0],
[ 1, 1, 1, 1, -1, -1, -1],
[ 2, 2, 2, 2, 2, -2, -2],
[ 3, 3, 3, 3, 3, 3, -3]])
One very short way to achieve what you want is to produce a meshgrid from your coordinates:
X,Y = np.meshgrid(x,y)
z = X+Y
or more general:
z = f(X,Y)
or even in one line:
z = f(*np.meshgrid(x,y))
EDIT:
If your function also may return a constant, you have to somehow infer the dimensions that the result should have. If you want to continue using meshgrids one very simple way would be re-write your function in this way:
def f(x,y):
return x*0+y*0+a
where a would be your constant. numpy would then take care of the dimensions for you. This is of course a bit weird looking, so instead you could write
def f(x,y):
return np.full(x.shape, a)
If you really want to go with functions that work both on scalars and arrays, it's probably best to go with np.vectorize as in #PaulPanzer's answer.
We have a Tensor of unknown length N, containing some int32 values.
How can we generate another Tensor that will contain N ranges concatenated together, each one between 0 and the int32 value from the original tensor ?
For example, if we have [4, 4, 5, 3, 1], the output Tensor should look like [0 1 2 3 0 1 2 3 0 1 2 3 4 0 1 2 0].
Thank you for any advice.
You can make this work with a tensor as input by using a tf.RaggedTensor which can contain dimensions of non-uniform length.
# Or any other N length tensor
tf_counts = tf.convert_to_tensor([4, 4, 5, 3, 1])
tf.print(tf_counts)
# [4 4 5 3 1]
# Create a ragged tensor, each row is a sequence of length tf_counts[i]
tf_ragged = tf.ragged.range(tf_counts)
tf.print(tf_ragged)
# <tf.RaggedTensor [[0, 1, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2], [0]]>
# Read values
tf.print(tf_ragged.flat_values, summarize=-1)
# [0 1 2 3 0 1 2 3 0 1 2 3 4 0 1 2 0]
For this 2-dimensional case the ragged tensor tf_ragged is a “matrix“ of rows with varying length:
[[0, 1, 2, 3],
[0, 1, 2, 3],
[0, 1, 2, 3, 4],
[0, 1, 2],
[0]]
Check tf.ragged.range for more options on how to create the sequences on each row: starts for inclusive lower limits, limits for exclusive upper limit, deltas for increment. Each may vary for each sequence.
Also mind that the dtype of the tf_counts tensor will propagate to the final values.
If you want to have everything as a tensorflow object, then use tf.range() along with tf.concat().
In [88]: vals = [4, 4, 5, 3, 1]
In [89]: tf_range = [tf.range(0, limit=item, dtype=tf.int32) for item in vals]
# concat all `tf_range` objects into a single tensor
In [90]: concatenated_tensor = tf.concat(tf_range, 0)
In [91]: concatenated_tensor.eval()
Out[91]: array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0], dtype=int32)
There're other approaches to do this as well. Here, I assume that you want a constant tensor but you can construct any tensor once you have the full range list.
First, we construct the full range list using a list comprehension, make a flat list out of it, and then construct a tensor.
In [78]: from itertools import chain
In [79]: vals = [4, 4, 5, 3, 1]
In [80]: range_list = list(chain(*[range(item) for item in vals]))
In [81]: range_list
Out[81]: [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0]
In [82]: const_tensor = tf.constant(range_list, dtype=tf.int32)
In [83]: const_tensor.eval()
Out[83]: array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0], dtype=int32)
On the other hand, we can also use tf.range() but then it returns an array when you evaluate it. So, you'd have to construct the list from the arrays and then make a flat list out of it and finally construct the tensor as in the following example.
list_of_arr = [tf.range(0, limit=item, dtype=tf.int32).eval() for item in vals]
range_list = list(chain(*[arr.tolist() for arr in list_of_arr]))
# output
[0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0]
const_tensor = tf.constant(range_list, dtype=tf.int32)
const_tensor.eval()
#output tensor as numpy array
array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0], dtype=int32)
How to place a list of numbers in to a 2D numpy array, where the second dimension of the array is equal to the number of digits of the largest number of that list? I also want the elements that don't belong to the original number to be zero in each row of the returning array.
Example:
From the list a = range(0,1001), how to get the numpy array of the below form:
[[0,0,0,0],
[0,0,0,1],
[0,0,0,2],
...
[0,9,9,8]
[0,9,9,9],
[1,0,0,0]]
Please note how the each number is placed in-place in a np.zeros((1000,4)) array at the end of the each row.
NB: A pythonic, vectorized implementation is expected
Broadcasting again!
def split_digits(a):
N = int(np.log10(np.max(a))+1) # No. of digits
r = 10**np.arange(N,-1,-1) # 10-powered range array
return (np.asarray(a)[:,None]%r[:-1])//r[1:]
Sample runs -
In [224]: a = range(0,1001)
In [225]: split_digits(a)
Out[225]:
array([[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 0, 0, 2],
...,
[0, 9, 9, 8],
[0, 9, 9, 9],
[1, 0, 0, 0]])
In [229]: a = np.random.randint(0,1000000,(7))
In [230]: a
Out[230]: array([431921, 871855, 636144, 541186, 410562, 89356, 476258])
In [231]: split_digits(a)
Out[231]:
array([[4, 3, 1, 9, 2, 1],
[8, 7, 1, 8, 5, 5],
[6, 3, 6, 1, 4, 4],
[5, 4, 1, 1, 8, 6],
[4, 1, 0, 5, 6, 2],
[0, 8, 9, 3, 5, 6],
[4, 7, 6, 2, 5, 8]])
Another concept using pandas str
def pir(a):
z = int(np.log10(np.max(a)))
s = pd.Series(a.astype(str))
zfilled = s.str.zfill(z + 1).sum()
a_ = np.array(list(zfilled)).reshape(-1, z + 1)
return a_.astype(int)
Using #Divakar's random array
a = np.random.randint(0,1000000,(7))
array([ 57190, 29950, 392317, 592062, 460333, 639794, 983647])
pir(a)
array([[0, 5, 7, 1, 9, 0],
[0, 2, 9, 9, 5, 0],
[3, 9, 2, 3, 1, 7],
[5, 9, 2, 0, 6, 2],
[4, 6, 0, 3, 3, 3],
[6, 3, 9, 7, 9, 4],
[9, 8, 3, 6, 4, 7]])