How to filter dataframe row based on length of column values

user_13331083 0
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I have a dataframe with one column containing the following strings:

df=pd.DataFrame(['Hello world', 'World is good', 'Worldisnice hello'], columns=['A'])

df
                     A
0         'Hello world'
1       'World is good'
2   'Worldisnice hello'

I'm trying to get the rows that contain one word with 11 characters of length

I'm using the following code by it is giving me the length of the string and not the word inside the column

df = df[df['A'].apply(lambda x: len(x) == 11)]

Getting the following result:

df
                     A
0         'Hello world'

And the output should be:

df
                     A
0   'Worldisnice hello'

Since is the only one that contains one word with length equals to 11 characters

Thank you

pandas dataframe python-3.x python
3 Answers

len(x) in your code is checking the len of the entire string.

>>> df.A.str.len()
 0    11
 1    13
 2    17

What you need to do instead is to split the string into words and check if any of the words' length is == 11.

The following code is what should do the job.

>>> df[df['A'].apply(lambda x: any(len(y) == 11 for y in x.split()))]
                  A
2  Worldisnice hello
user_64150800

Another approach:

df[df.A.str.split().map(lambda x: any(len(y) == 11 for y in x))]

that provides:

                   A
2  Worldisnice hello
user_53597970

I like to explicitly define simple filtering functions. I find it way more readable and maintainable.

In [8]: def f(row):
   ...:     words = row.A.split()
   ...:     for w in words:
   ...:         if len(w) == 11:
   ...:             return True
   ...: 

In [9]: df.loc[df.apply(f, axis=1) == True]
Out[9]: 
                   A
2  Worldisnice hello
user_31900760
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