Replacing a part of string in a Dataframe

Question

I want to replace a part of string in a dataframe. For example,

df =
index                    group
2019-02-01 09:30:07     8157_S1
2019-02-01 09:30:23    S_1.M_01
2019-02-01 09:30:38    S_1.M_02
2019-02-01 09:32:11    S_1.M_08
2019-02-01 09:32:27    S_1.M_12
2019-02-01 09:32:42    S_1.M_11
2019-02-01 09:32:57    S_1.M_10
2019-02-01 09:33:13    S_1.M_09
2019-02-01 09:33:30     8157_S2
2019-02-01 09:33:47    S_2.M_13

I want to remove "8157_" and "S_1." and "S_2.". My present code is

df['group'] = df['group'].replace("8157_", "") 
df['group'] = df['group'].replace("S_1.","")

There was no change in the output.

But my expected output is

df =
index                    group
2019-02-01 09:30:07    S1
2019-02-01 09:30:23    M_01
2019-02-01 09:30:38    M_02
2019-02-01 09:32:11    M_08
2019-02-01 09:32:27    M_12
2019-02-01 09:32:42    M_11
2019-02-01 09:32:57    M_10
2019-02-01 09:33:13    M_09
2019-02-01 09:33:30     S2
2019-02-01 09:33:47    M_13

use df['group'].str.replace

sammywemmy
– sammywemmy

2020-01-22 00:53:24 +00:00
Commented Jan 22, 2020 at 0:53 — sammywemmy
– sammywemmy, Commented Jan 22, 2020 at 0:53
Or add regex=True to your replace

Chris
– Chris

2020-01-22 00:53:55 +00:00
Commented Jan 22, 2020 at 0:53 — Chris
– Chris, Commented Jan 22, 2020 at 0:53

Chris · Accepted Answer · 2020-01-22 00:55:19Z

1

Using pandas.Series.str.replace:

df['group'] = df['group'].str.replace("S_[12]\.|8157_","") 
print(df)

Output:

                 index group
0  2019-02-01 09:30:07    S1
1  2019-02-01 09:30:23  M_01
2  2019-02-01 09:30:38  M_02
3  2019-02-01 09:32:11  M_08
4  2019-02-01 09:32:27  M_12
5  2019-02-01 09:32:42  M_11
6  2019-02-01 09:32:57  M_10
7  2019-02-01 09:33:13  M_09
8  2019-02-01 09:33:30    S2
9  2019-02-01 09:33:47  M_13

answered Jan 22, 2020 at 0:55

Chris

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2 Comments

Mainland Over a year ago

Just great! your one line code did exactly what I wanted. In between, S_[12]\. means what?

Chris Over a year ago

It means replace S_{1 or 2}\.. Note the backslash for \.; without it, . is interpreted as any char which is not in your case.

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