pandas replace multiple values one column

Question

In a column risklevels I want to replace Small with 1, Medium with 5 and High with 15. I tried:

dfm.replace({'risk':{'Small': '1'}},
            {'risk':{'Medium': '5'}},
            {'risk':{'High': '15'}})

But only the medium were replaced. What is wrong ?

Jeff · Accepted Answer · 2014-02-28 16:24:50Z

93

Your replace format is off

In [21]: df = pd.DataFrame({'a':['Small', 'Medium', 'High']})

In [22]: df
Out[22]: 
        a
0   Small
1  Medium
2    High

[3 rows x 1 columns]

In [23]: df.replace({'a' : { 'Medium' : 2, 'Small' : 1, 'High' : 3 }})
Out[23]: 
   a
0  1
1  2
2  3

[3 rows x 1 columns]

answered Feb 28, 2014 at 16:24

Jeff

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2 Comments

EdChum Over a year ago

I wasn't sure what was wrong with the replace format line so I suggested using map instead. +1 for spotting OP error

Simone Over a year ago

This solution only works in the current example when the cell in the dataframe contains one word. If there is more than one word re.sub() worked for me. Used single commands / code lines. I am sure there is a fancier solution (maybe with translate ?) But it's a time management question...

Surya Chhetri · Accepted Answer · 2017-03-15 21:26:24Z

In [123]: import pandas as pd                                                                                                                                

In [124]: state_df = pd.DataFrame({'state':['Small', 'Medium', 'High', 'Small', 'High']})                                                                    

In [125]: state_df
Out[125]: 
    state
0   Small
1  Medium
2    High
3   Small
4    High

In [126]: replace_values = {'Small' : 1, 'Medium' : 2, 'High' : 3 }                                                                                          

In [127]: state_df = state_df.replace({"state": replace_values})                                                                                             

In [128]: state_df
Out[128]: 
   state
0      1
1      2
2      3
3      1
4      3

EdChum · Accepted Answer · 2014-02-28 16:19:44Z

12

You could define a dict and call map

In [256]:

df = pd.DataFrame({'a':['Small', 'Medium', 'High']})
df
Out[256]:
        a
0   Small
1  Medium
2    High

[3 rows x 1 columns]
In [258]:

vals_to_replace = {'Small':'1', 'Medium':'5', 'High':'15'}
df['a'] = df['a'].map(vals_to_replace)
df
Out[258]:
    a
0   1
1   5
2  15

[3 rows x 1 columns]


In [279]:

val1 = [1,5,15]
df['risk'].update(pd.Series(val1))
df
Out[279]:
  risk
0    1
1    5
2   15

[3 rows x 1 columns]

edited Feb 28, 2014 at 16:19

answered Feb 28, 2014 at 16:11

EdChum

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2 Comments

EdChum Over a year ago

@Jeff not familiar with that method, am I using it correctly?

Jeff Over a year ago

yes that is correct (but I realize that the issue is that the OP replace format is wrong)

buhtz · Accepted Answer · 2021-08-06 11:36:32Z

9

Looks like OP may have been looking for a one-liner to solve this through consecutive calls to .str.replace:

dfm.column = dfm.column.str.replace('Small', '1') \
    .str.replace('Medium', '5') \
        .str.replace('High', '15')

OP, you were close but just needed to replace your commas with .str.replace and the column call ('risk') in a dictionary format isn't necessary. Just pass the pattern-to-match and replacement-value as arguments to replace.

edited Aug 6, 2021 at 11:36

buhtz

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answered Aug 31, 2018 at 16:44

ChrisDanger

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3 Comments

Mark Dickinson Over a year ago

Welcome to Stack Overflow. Please could you add some explanation to your answer? (What changes did you make, and why? Why did the OPs original code not work?) Without the explanation, the answer isn't so useful to future visitors.

NickD Over a year ago

What does this answer add that the other answers lack?

Jeff D. White Over a year ago

This answer is useful if you want to replace a piece of the string and not the entire string. I found this answer when trying to understand if you could put together multiple .str.replace in a statement. That said, if the match is desired to the total string (OPs question), and not a piece of the string, the preferred answer is best.

Amir Pourmand · Accepted Answer · 2021-07-05 23:41:49Z

7

I had to turn on the "regex" flag to make it work:

 df.replace({'a' : {'Medium':2, 'Small':1, 'High':3 }}, regex=True)

edited Jul 5, 2021 at 23:41

Amir Pourmand

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answered Apr 13, 2020 at 13:48

Mehdi Rostami

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Comments

Antonio · Accepted Answer · 2017-09-25 19:38:39Z

2

String replace each string (Small, Medium, High) for the new string (1,5,15)\

If dfm is the dataframe name, column is the column name.

dfm.column = dfm.column.str.replace('Small', '1')
dfm.column = dfm.column.str.replace('Medium', '5')
dfm.column = dfm.column.str.replace('High', '15')

answered Sep 25, 2017 at 19:38

Antonio

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Comments

markling · Accepted Answer · 2022-08-07 11:41:58Z

2

Use series.replace with lists of before and after values for greater ease:

df.risklevels = df.risklevels.replace( ['Small','Medium','High'], [1,2,3] )

See here.

answered Aug 7, 2022 at 11:41

markling

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