Printing FULL contents of numpy array [duplicate] - python

This question already has answers here:
How do I print the full NumPy array, without truncation?
(22 answers)
Closed 9 years ago.
I am working with image processing in python and I want to output a variable, right now the variable b is a numpy array with shape (200,200). When I do print b all I see is:
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.]])
How do I print out the full contents of this array, write it to a file or something simple so I can just look at the contents in full?

Of course, you can change the print threshold of the array as answered elsewhere with:
np.set_printoptions(threshold=np.nan)
But depending on what you're trying to look at, there's probably a better way to do that. For example, if your array truly is mostly zeros as you've shown, and you want to check whether it has values that are nonzero, you might look at things like:
import numpy as np
import matplotlib.pyplot as plt
In [1]: a = np.zeros((100,100))
In [2]: a
Out[2]:
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.]])
Change some values:
In [3]: a[4:19,5:20] = 1
And it still looks the same:
In [4]: a
Out[4]:
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.]])
Check some things that don't require manually looking at all values:
In [5]: a.sum()
Out[5]: 225.0
In [6]: a.mean()
Out[6]: 0.022499999999999999
Or plot it:
In [7]: plt.imshow(a)
Out[7]: <matplotlib.image.AxesImage at 0x1043d4b50>
Or save to a file:
In [11]: np.savetxt('file.txt', a)

to_print = "\n".join([", ".join(row) for row in b])
print (to_print) #console
f = open("path-to-file", "w")
f.write(to_print) #to file
In case it's numpy array: Print the full numpy array

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How to store a numpy array with shape (32,32) as (1,32,32)

For various reasons, I need to transfer a 2d numpy array of shape (32,32) to 3d shape (1,32,32). Is there a way to do this?
Just use np.reshape:
>>> a = np.zeros((32, 32))
>>> a.reshape((1, 32, 32))
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.]]])
>>>
I used a array of zeros for demonstrating.

Replace values in bigger numpy array with smaller array

I have 2 numpy arrays: The bigger one is a 10 x 10 numpy array and the smaller one is a 2 x 2 array.
I would like to substitute the values in the bigger array with those from the smaller array, at a user specified location. E.g. Replace the values of the 10 x 10 array starting from its center point by replacing 4 values with the 2 x 2 array.
Right now, I am doing this by using a nested for loop, and figuring out which pixels in the bigger array overlap those of the smaller array. Is there a more pythonic way to do it?
In [1]: import numpy as np
In [2]: a = np.zeros(100).reshape(10,10)
In [3]: b = np.ones(4).reshape(2,2)
In [4]: a[4:6, 4:6] = b
In [5]: a
Out[5]:
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., 1., 1., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 1., 1., 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.]])

Python: two questions about a "numpy.ndarray"

I created a numpy.ndarray with value
import numpy as np
from numpy import nonzero
data = np.zeros((5, 5))
data
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.]])
i wish to change some values with 1
data[0,0] = 1
data[4,4] = 1
data
array([[ 1., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 1.]])
if i change 0 with 5 using negative values i have
data[-5,-5] = 5
data[-4,-4] = 5
>>> data
array([[ 5., 0., 0., 0., 0.],
[ 0., 5., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.]])
1- I don't understand why i have not an error message as
>>> data[10,10] = 5
Traceback (most recent call last):
File "<interactive input>", line 1, in <module>
IndexError: index (10) out of range (0<=index<5) in dimension 0
2- it's not clear why with data[-5,-5] = 5 and data[-4,-4] = 5 the value 5 is inserted in position 0,0 and 1,1
From the documentation:
Negative indices are interpreted as counting from the end of the array
This is the standard Python indexing behavior (used in Python lists, etc.).

Using numpy to Obtain Submatrix

I'm trying to do the following with numpy (python newbie here)
Create a zeroed matrix of the rigth dimensions
num_rows = 80
num_cols = 23
A = numpy.zeros(shape=(num_rows, num_cols))
Operate on the matrix
k = 5
numpy.transpose(A)
U,s,V = linalg.svd(A)
Extract sub-matrix
sk = s[0:(k-1), 0:(k-1)]
Results on error
Traceback (most recent call last):
File "tdm2svd.py", line 40, in <module>
sk = s[0:(k-1), 0:(k-1)]
IndexError: too many indices
What am I doing wrong?
to answer your question s is only a 1d array ... (even if you did actually transpose it ... which you did not)
>>> u,s,v = linalg.svd(A)
>>> s
array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
>>>
for selecting a submatrix
I think this does what you want ... there may be a better way
>>> rows = range(10,15)
>>> cols = range(5,8)
>>> A[rows][:,cols]
array([[ 0., 0., 0.],
[ 0., 0., 0.],
[ 0., 0., 0.],
[ 0., 0., 0.],
[ 0., 0., 0.]])
or probably better
>>> A[15:32, 2:7]
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.]])

Python numpy array strange return

I am relatively new to python. I have a numpy array that has 3 dimensions. I know we can display only few elements using :.
It seems to work just fine while I'm doing it starting from a small value, but at one point, it returns something different than a matrix.
I want to get the mean value for the array. So, for instance, given an array c, I do numpy.mean(c[0:200][0:200][0:200]). This works just fine. But increasing the starting point (i.e. c[200:][200:][200:]) doesn't work and returns nan. So, printing the result explains the nan value. But I don't get why c[200:][200:][200:] returns this kind of answer.
Here's two examples:
In [68]: c.shape
Out[68]: (448, 433, 446)
In [63]: c[100:][100:][100:]
Out[63]:
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.]]])
In [67]: c[200:][200:][200:]
Out[67]: array([], shape=(0, 433, 446), dtype=float64)
You're indexing into the arrays improperly. The way to index on multiple dimensions is array[x, y, z], not array[x][y][z]. So you want to do c[200:, 200:, 200:].
When you use a single index in brackets, it indexes into the first dimension. So when you do c[200:][200:][200:], you try to get the first 200 elements of the array along the first dimension every time. But that dimension is less than 600 elements long, so when you do it three times there's nothing left to get.

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