CODEWITHSUNDEEP
regexpython-3.xpandasreplace
mikeqfu
mikeqfu

Reputation: 328

pandas Series replace using dictionary with regex keys

Suppose there is a dataframe defined as

df = pd.DataFrame({'Col_1': ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', '0'], 
                   'Col_2': ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', '0']})

which looks like

   Col_1 Col_2
0      A     a
1      B     b
2      C     c
3      D     d
4      E     e
5      F     f
6      G     g
7      H     h
8      I     i
9      J     j
10     0     0

I would like to replace the values in Col_1 by using a dictionary defined as

repl_dict = {re.compile('[ABH-LP-Z]'): 'DDD',
             re.compile('[CDEFG]'): 'BBB WTT',
             re.compile('[MNO]'): 'AAA WTT',
             re.compile('[0-9]'): 'CCC'}

I would expect to get a new dataframe in which the Col_1 should have been as follows

      Col_1
0       DDD
1       DDD
2   BBB WTT
3   BBB WTT
4   BBB WTT
5   BBB WTT
6   BBB WTT
7       DDD
8       DDD
9       DDD
10      CCC

I just simply use df['Col_1'].replace(repl_dict, regex=True). However, it does not produce what I expected. What I've got is like:

                      Col_1
0     BBB WTTBBB WTTBBB WTT
1     BBB WTTBBB WTTBBB WTT
2                   BBB WTT
3                   BBB WTT
4                   BBB WTT
5                   BBB WTT
6                   BBB WTT
7     BBB WTTBBB WTTBBB WTT
8     BBB WTTBBB WTTBBB WTT
9     BBB WTTBBB WTTBBB WTT
10                      CCC

I would appreciate it very much if anyone could let me know why the df.replace() was not working for me and what would be a correct way to replace multiple values to get the expected output.

Upvotes: 1

Views: 4315

Answers (2)

TejasKhajanchee
TejasKhajanchee

Reputation: 103

A more realistic scenario could be where you would want reclassify entries based on a pattern as follows:

Consider dataframe 'x' as follows:

             column
0       good farmer
1        bad farmer
2         ok farmer
3  worker did wrong
4      worker fired
5      worker hired
6   heavy duty work
7   light duty work

Then consider the following code:

x['column_reclassified'] = x['column'].replace(
    to_replace=[
        '^.*(farmer).*$',
        '^.*(worker).*$',
        '^.*(duty).*$'
    ],
    value=[
        'FARMER',
        'WORKER',
        'DUTY'
    ],
    regex=True
)

and it will produce the following output:

             column column_reclassified
0       good farmer              FARMER
1        bad farmer              FARMER
2         ok farmer              FARMER
3  worker did wrong              WORKER
4      worker fired              WORKER
5      worker hired              WORKER
6   heavy duty work                DUTY
7   light duty work                DUTY

Hope this also helps.

Upvotes: 0

Jan
Jan

Reputation: 43169

Use anchors (^ and $, that is):

repl_dict = {re.compile('^[ABH-LP-Z]$'): 'DDD',
             re.compile('^[CDEFG]$'): 'BBB WTT',
             re.compile('^[MNO]$'): 'AAA WTT',
             re.compile('^[0-9]+$'): 'CCC'}

Which produces with df['Col_1'].replace(repl_dict, regex=True):

0         DDD
1         DDD
2     BBB WTT
3     BBB WTT
4     BBB WTT
5     BBB WTT
6     BBB WTT
7         DDD
8         DDD
9         DDD
10        CCC

Upvotes: 3

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