Count number of 1's in an array in Python [duplicate] - python

This question already has answers here:
How do I count the occurrence of a certain item in an ndarray?
(32 answers)
Closed 14 days ago.
I have an array A. I want to count the number of 1's in the array but I am getting an error. I present the expected output.
import numpy as np
A=np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0,
0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
B=np.count(A==1)
print(B)
The error is
in __getattr__
raise AttributeError("module {!r} has no attribute "
AttributeError: module 'numpy' has no attribute 'count'
The expected output is
7


The np.count function does not exist, you can use np.count_nonzero instead.
import numpy as np
A=np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0,
0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
B=np.count_nonzero(A==1) # or just A, if you'll never have anything other than 0 or 1
print(B)

Related

Count occurrences of a value in a matrix [closed]

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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 months ago.
Improve this question
I have a problem with my code:
I don't manage to find a right way to count occurrences of number one in a list of arrays like this
[[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1],
[1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1], [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1]]
I want to produce an array that contains the sum of ones in each list of my matrix
I tried to use numpy making
for m in matrix:
y = m.np
y.count(1)
but the compiler returns me a message like this:
module 'numpy' has no attribute 'm'
How can I proceed?

Since you are counting ones, you can just sum them up:
count_ones = [sum(array) for array in matrix]

Grow a list with the counts; there is no need of numpy array.
ones_per_row = []
for m in matrix:
ones_per_row.append(m.count(1))
print(ones_per_row)
Output:
[7, 2, 2, 7, 8, 3, 7, 3]

Indexing 2 dimensional array in python

I've been trying to change a single item in a 2-dimensional array in python using the syntax x[2][3]=1 but instead of just changing the item in the 2nd row 3rd column, it ends up changing the values of all of the 3rd column. My code is below:
population = [[0]*20]*5
population[2][3] = 1
for row in population:
print(row)
This outputs
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
but I only want
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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 would I index the item such that it only changes the 2nd row and 3rd column?
I'm using python 3.7.4 on repl.it
Link here: https://repl.it/#ajqe/2d-array-test

Use :
population = [[0]*20 for _ in range(5)]
to generate the lists instead. The method you are using is referencing the same object 5 times, instead of creating 5 separate lists. To check this you can use the is operator:
>>> population = [[0]*20]*5
>>> population[0] is population[1]
True
>>> population = [[0]*20 for _ in range(5)]
>>> population[0] is population[1]
False

python: convert array of integers to array of their binary representation [duplicate]

I need a way to convert 20 million 32 and 64-bit integers into corresponding bit arrays (so this has to be memory/time efficient). Under advice from a different question/answer here on SO, I'm attempting to do this by using numpy.unpackbits. While experimenting with this method I ran into unexpected results:
np.unpackbits(np.array([1], dtype=np.uint64).view(np.uint8))
produces:
array([0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8)
I would expect the 1 element to be the last one, but not in the middle. So I'm obviously missing something that preserves the byte order. What am I missing?

Try: dtype='>i8', like so:
In [6]: np.unpackbits(np.array([1], dtype='>i8').view(np.uint8))
Out[6]:
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, 1], dtype=uint8)
Reference:
http://docs.scipy.org/doc/numpy/user/basics.byteswapping.html

How to manipulate a pandas series with an alternating logic without using for loop?

I have a series comprising only of 1, 0 and -1. I am trying to implement a flipflop logic : 1 can come only after -1 and -1 can come only after a 1. So basically, 1 and -1 have to occur only after they have already occurred.
>>> a
[0, 0, 1, 1, 0, 0, 0, 0, -1, 0, 0, 0, -1, -1, -1, -1, 0, 0, 0, 1, 1, 0, 0, -1, 1]
>>> out
[0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 1]
How can I achieve this using pandas and without using a for loop?
Please help!

I have taken your list in a panda series to utilize pandas functionality
a = pd.Series([0, 0, 1, 1, 0, 0, 0, 0, -1, 0, 0, 0, -1, -1, -1, -1, 0, 0, 0, 1, 1, 0, 0, 1, 1])
after running the below code it gives the desire output
a[a.index.difference(a[a!=0][a[a!=0].diff()!=0].index)] = 0
output
[0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0]
I hope it would give you the desired solution.

using numpy to convert an int to an array of bits

I need a way to convert 20 million 32 and 64-bit integers into corresponding bit arrays (so this has to be memory/time efficient). Under advice from a different question/answer here on SO, I'm attempting to do this by using numpy.unpackbits. While experimenting with this method I ran into unexpected results:
np.unpackbits(np.array([1], dtype=np.uint64).view(np.uint8))
produces:
array([0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8)
I would expect the 1 element to be the last one, but not in the middle. So I'm obviously missing something that preserves the byte order. What am I missing?

Try: dtype='>i8', like so:
In [6]: np.unpackbits(np.array([1], dtype='>i8').view(np.uint8))
Out[6]:
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, 1], dtype=uint8)
Reference:
http://docs.scipy.org/doc/numpy/user/basics.byteswapping.html

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