BUG: Pandas squashes 1-dimensional Numpy array with shape (1,) down to a 0-dimensional array about pandas HOT 7 OPEN

can I take this issue ?

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take

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@sunghjung3 I think it seems to be a regression in recent versions of pandas. It appears that a 1-dimensional numpy array with a single entry is being automatically converted to a 0-dimensional array when stored in a DataFrame.
Temporary Workaround:

1. Until the issue is resolved in future pandas releases, you can explicitly check and reshape the arrays as needed.

import pandas as pd
import numpy as np

df = pd.DataFrame(columns=['x'])
arr0 = np.array([1., 2.])
df.loc[0, "x"] = arr0
print(df.loc[0, "x"].shape)  # prints (2,)

arr1 = np.array([1.])
df.loc[1, "x"] = arr1 if arr1.shape != () else arr1.reshape(1,)
print(df.loc[1, "x"].shape)  # prints (1,)

Hope this helps,
Thanks

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@Siddharth-Latthe-07 I've already tried this, and it doesn't work. It still prints (). Please try it yourself.

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oh, I see then can we create a custom function to ensure that the shape of the numpy array is preserved when you store it in the DataFrame. This function will explicitly check the shape of the array and reshape it if necessary before storing it in the DataFrame.
Thoughts ? @sunghjung3

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something like this:-

def safe_loc_assign(df, index, column, value):
    if isinstance(value, np.ndarray) and value.shape == ():
        value = value.reshape(1,)
    df.at[index, column] = value

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@Siddharth-Latthe-07 I don't get the logic of your function.

I'm assuming that we would call this function like safe_loc_assign(df, 1, "x", arr1). And inside this function, it checks for value.shape == (). But the shape of the value argument (which is arr1) is not (), so this if statement isn't even utilized.

And the result is still the same; it prints ().

Also, even if I change this if-statement to value.shape == (1,), it still doesn't work.

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