Home/ Questions/Quick handling of the ''can only use .str accessor with string values, which use np.object_ dtype in pandas'' error.
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Preston Li
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Asked: 2022-05-18T21:06:16+00:00 In:

Quick handling of the ”can only use .str accessor with string values, which use np.object_ dtype in pandas” error.

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Hi everyone, I’m learning about python. While working, I try undefined. As a result, I get the message:

dc_listings['price'].str.replace(',', '')
 AttributeError: Can only use .str accessor with string values, which use np.object_ dtype in pandas

What can I do about the “can only use .str accessor with string values, which use np.object_ dtype in pandas” issue? Is there a better approach?

use .str accessor with string valueswhich use np.object_ dtype
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2 Answers

  1. Best Answer
    Expert

    The cause:

    You get this error because you are trying to change a pattern in a string column, but the column that you are working with is not really a string.

    Solution:

    You can only use .str with string columns, and you have a float64, as the error shows. What you have will not  really do anything because a float won’t contain any commas, but generally speaking, you can ignore it first:

    dc_listings['price'].astype(str).str.replace...

    Let’s consider the following suggestion to solve your problem:

    In [18]: df
Out[18]:
a b c d e
0 0.645821 0.152197 0.006956 0.600317 0.239679
1 0.865723 0.176842 0.226092 0.416990 0.290406
2 0.046243 0.931584 0.020109 0.374653 0.631048
3 0.544111 0.967388 0.526613 0.794931 0.066736
4 0.528742 0.670885 0.998077 0.293623 0.351879

In [19]: df['a'].astype(str).str.replace("5", " hi ")
Out[19]:
0 0.64 hi 8208 hi hi 4779467
1 0.86 hi 7231174332336
2 0.04624337481411367
3 0. hi 44111244991 hi 194
4 0. hi 287421814241892
Name: a, dtype: object
    • 0

  2. There are two ways to do it:

    1. This error can be fixed with series. 

      dc_listings['price'].series.str.replace(',', '')

    1. If series fails to work, you can use apply(str) instead. 

      dc_listings['price'].apply(str).str.replace(',', '')
    • 7

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