Why does `int in List[List[int]]` return `False` but `np.int in List[List[int]]` return `True`?-CodePudding

I observed this strange behavior while checking if a number is in a list. If the number is of generic int type, the check failed; but the check went through successfully if the number is of numpy.int64 type. Can anyone explain why? I know I could do better by generating the list lst=df['A'].values.tolist() to get a list of integers instead of a list of list. But my question is why the numpy.int64 would work below?

import pandas as pd
import numpy as np

df = pd.DataFrame({'A': range(31, 36)})
print(df)

#     A
# 0  31
# 1  32
# 2  33
# 3  34
# 4  35

lst=df.values.tolist()
print(lst)
# [[31], [32], [33], [34], [35]]

x=31
print(x) # 31
print(type(x)) # <class 'int'>
if x in lst:
    print('Yes')
else:
    print('No')

# prints No!

y=df['A'][0]
print(y)  # 31
print(type(y)) # <class 'numpy.int64'>
if y in lst:
    print('Yes')
else:
    print('No')

# prints Yes

CodePudding user response：

Your list doesn't contain 31. It contains another list that contains 31, but it doesn't directly contain 31.

thing in lst works like this:

for x in lst:
    if x is thing or x == thing:
        return True
return False

When you check whether a regular int is in your list, x == thing is always False, because all elements of your list are more lists, and an int is never equal to a list. However, with a numpy.int64, the comparison broadcasts. When you compare

numpy.int64(31) == [31]

[31] is converted to a NumPy array, and you get an array of elementwise comparison results, comparing numpy.int64(31) to every element of the array, resulting in

numpy.array([True])

A one-element NumPy boolean array is treated as its single element in an if check, so when the list's in logic compares numpy.int64(31) against [31], it thinks these are equal, and it reports True as the result of the in check.