CODEWITHSUNDEEP
pythonnumpy
user
user

Reputation: 79

Numpy error: operands could not be broadcast together with shapes 1

I am a python beginner. I have an error message "ValueError: operands could not be broadcast together with shapes".

Here is my data:

import numpy as np
spent = np.array([
    10,  10,  13,  12,  109,   17,  31, 1,   39, 41,   45,
    41, 71,   161,   39,  115,    5,   51,   58,  334,  165, 1032,
    40,  52, 21,   68,  79,  482,  10,  265,  60,  67,   12,
    53,  188,  32,  397,  51, 17, 156,  100,  85,  53,  95,
     68,  308,   53,  675,   78,  27,  219,  45,  45,   30,   61,
    16,   72,   80,   96, 1386,  370,   16,   81,   28,   43,   90,
    33,   66,   77])
visit = np.array([
   19, 13, 16, 16, 18, 9, 12, 3, 15, 16, 16, 3, 4, 11, 11, 11, 11,
   12, 12, 12, 13, 13, 14, 14, 15, 15, 5, 6, 6, 7, 7, 7, 7, 7,
   17, 17, 8, 8, 8, 4, 4, 13, 8, 4, 4, 9, 20, 10, 11, 11, 14,
   12, 12, 15, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 16, 16, 18, 11,
   6])

My job is select spent>100 and visit>10 together. So, I would like to find people who pay more than $100 among people who visited more than 10 times. I have tried the following codes.

a=spent[spent>100] & [visit>10]
print(a)

But, I have an error message "ValueError: operands could not be broadcast together with shapes". Could you advise me how to deal with this? I just have no idea.

Upvotes: 1

Views: 2427

Answers (2)

EdChum
EdChum

Reputation: 394449

IIUC you don't need the mask on spent per se like you've done:

In[16]:
a=(spent>100) & (visit>10)
a

Out[16]: 
array([False, False, False, False,  True, False, False, False, False,
       False, False, False, False,  True, False,  True, False, False,
       False,  True,  True,  True, False, False, False, False, False,
       False, False, False, False, False, False, False,  True, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False,  True, False, False, False,
       False, False, False, False, False,  True,  True, False, False,
       False, False, False, False, False, False], dtype=bool)

This gives you a boolean mask that is only True where both conditions are met in both arrays, you can then use this to mask against the original arrays

So using this against spent:

In[18]:
spent[a]

Out[18]: array([ 109,  161,  115,  334,  165, 1032,  188,  219, 1386,  370])

Your error was that you masked your original array which produced an array that was a different shape that what you're trying to broadcast against visit:

print(spent[spent>100].shape)
print((visit>10).shape)
(16,)
(69,)

You could compound the conditions into the same mask:

In[20]:
spent[(spent > 100) & (visit > 10)]

Out[20]: array([ 109,  161,  115,  334,  165, 1032,  188,  219, 1386,  370])

to produce the same result

Upvotes: 3

Carles Mitjans
Carles Mitjans

Reputation: 4866

A possible solution would be using list comprehensions:

[(x, y) for x, y in zip(visit, spent) if x > 10 and y > 100]

You could also use numpy as follows:

spent[visit > 10] > 100

Upvotes: 0

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