I have a hard time doing the arithmetic statement

Convert the tuple, "Third" to an array.

Double all the elements in "Third"

Add the elements in this array to the elements in "Fourth".

To do this, use only 1 arithmetic statement. Store this result as "Fifth" and print it.

Hint Answer should look like this:

Fifth = 2*(_) _____ print (Fifth)

third = (-6,-18,-10,-4,-90,-55,-56)
fourth =[6],[18],[10],[4],[90],[55],[56]
fourth = np.array(fourth)
print(third)
print(type(third))
print("After converting Python tuple to array")

arr = np.asarray(third)
print(arr)
print(type(arr))

fifth = 2*arr
print(fifth)

output:
(-6, -18, -10, -4, -90, -55, -56)
<class 'tuple'>
After converting Python tuple to array
[ -6 -18 -10  -4 -90 -55 -56]
<class 'numpy.ndarray'>

[ -12  -36  -20   -8 -180 -110 -112]

CodePudding user response：

The numpy.ndarray type (docs) supports addition using the operator. However, in your example my guess is that 2*arr fourth will not produce the result you are looking for because fourth is of shape (7,1) while arr is of shape (7,), so arr fourth would be of shape (7,7).

There are multiple ways to get fourth into shape (7,). fourth.flatten() will work (docs), as will numpy indexing, e.g. fourth[:,0].

Try this runnable (and editable!) example

import numpy as np

third = (-6,-18,-10,-4,-90,-55,-56)
fourth =[6],[18],[10],[4],[90],[55],[56]
fourth = np.array(fourth)
print(third)
print(type(third))
print("After converting Python tuple to array")

arr = np.asarray(third)
print(arr)
print(type(arr))

print(f"{arr.shape=}, {fourth.shape=}, {fourth.flatten().shape=}, {fourth[:,0].shape=}")

fifth = 2*arr   fourth[:,0]
print(f"{fifth=}")

fifth

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