CODEWITHSUNDEEP
pythonpython-3.xnumpy
Divyanshu
Divyanshu

Reputation: 363

Numpy - Different behavior for 1-d and 2-d arrays

I was reviewing some numpy code and came across this issue. numpy is exhibiting different behavior for 1-d array and 2-d array. In the first case, it is creating a reference while in the second, it is creating a deep copy.

Here's the code snippet

import numpy as np

# Case 1: when using 1d-array

arr = np.array([1,2,3,4,5])
slice_arr = arr[:3]  # taking first three elements, behaving like reference

slice_arr[2] = 100 # modifying the value

print(slice_arr)
print (arr) # here also value gets changed

# Case 2: when using 2d-array

arr = np.array([[1,2,3],[4,5,6],[7,8,9]])
slice_arr = arr[:,[0,1]]  # taking all rows and first two columns, behaving like deep copy

slice_arr[0,1] = 100 # modifying the value

print(slice_arr)
print() # newline for clarity
print (arr) # here value doesn't change

Can anybody explain the reason for this behavior?

Upvotes: 1

Views: 76

Answers (1)

Matthieu Brucher
Matthieu Brucher

Reputation: 22033

The reason is that you are not slicing in the same way, it's not about 1D vs 2D.

slice_arr = arr[:3]

Here you are using the slicing operator, so numpy can make a view on your original data and returns it.

slice_arr = arr[:,[0,1]]

Here you are using a list of elements you want, and it's not a slice (even if it could be represented by a slice), in that case, numpy returns a copy.

All these are getters, so they can return either view or copy.

For setters, it's always modifying the current array.

Upvotes: 4

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