Remove unwanted parts from strings in a column

Question:

I am looking for an efficient way to remove unwanted parts from strings in a DataFrame column.

Data looks like:

    time    result
1    09:00   +52A
2    10:00   +62B
3    11:00   +44a
4    12:00   +30b
5    13:00   -110a

I need to trim these data to:

    time    result
1    09:00   52
2    10:00   62
3    11:00   44
4    12:00   30
5    13:00   110

I tried .str.lstrip('+-') and .str.rstrip('aAbBcC'), but got an error:

TypeError: wrapper() takes exactly 1 argument (2 given)

Any pointers would be greatly appreciated!

Asked By: Yannan Wang

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Answers:

data['result'] = data['result'].map(lambda x: x.lstrip('+-').rstrip('aAbBcC'))
Answered By: eumiro

There’s a bug here: currently cannot pass arguments to str.lstrip and str.rstrip:

http://github.com/pydata/pandas/issues/2411

EDIT: 2012-12-07 this works now on the dev branch:

In [8]: df['result'].str.lstrip('+-').str.rstrip('aAbBcC')
Out[8]: 
1     52
2     62
3     44
4     30
5    110
Name: result
Answered By: Wes McKinney

In the particular case where you know the number of positions that you want to remove from the dataframe column, you can use string indexing inside a lambda function to get rid of that parts:

Last character:

data['result'] = data['result'].map(lambda x: str(x)[:-1])

First two characters:

data['result'] = data['result'].map(lambda x: str(x)[2:])
Answered By: prl900

i’d use the pandas replace function, very simple and powerful as you can use regex. Below i’m using the regex D to remove any non-digit characters but obviously you could get quite creative with regex.

data['result'].replace(regex=True,inplace=True,to_replace=r'D',value=r'')
Answered By: Coder375

I often use list comprehensions for these types of tasks because they’re often faster.

There can be big differences in performance between the various methods for doing things like this (i.e. modifying every element of a series within a DataFrame). Often a list comprehension can be fastest – see code race below for this task:

import pandas as pd
#Map
data = pd.DataFrame({'time':['09:00','10:00','11:00','12:00','13:00'], 'result':['+52A','+62B','+44a','+30b','-110a']})
%timeit data['result'] = data['result'].map(lambda x: x.lstrip('+-').rstrip('aAbBcC'))
10000 loops, best of 3: 187 µs per loop
#List comprehension
data = pd.DataFrame({'time':['09:00','10:00','11:00','12:00','13:00'], 'result':['+52A','+62B','+44a','+30b','-110a']})
%timeit data['result'] = [x.lstrip('+-').rstrip('aAbBcC') for x in data['result']]
10000 loops, best of 3: 117 µs per loop
#.str
data = pd.DataFrame({'time':['09:00','10:00','11:00','12:00','13:00'], 'result':['+52A','+62B','+44a','+30b','-110a']})
%timeit data['result'] = data['result'].str.lstrip('+-').str.rstrip('aAbBcC')
1000 loops, best of 3: 336 µs per loop
Answered By: tim654321

A very simple method would be to use the extract method to select all the digits. Simply supply it the regular expression 'd+' which extracts any number of digits.

df['result'] = df.result.str.extract(r'(d+)', expand=True).astype(int)
df

    time  result
1  09:00      52
2  10:00      62
3  11:00      44
4  12:00      30
5  13:00     110
Answered By: Ted Petrou

How do I remove unwanted parts from strings in a column?

6 years after the original question was posted, pandas now has a good number of “vectorised” string functions that can succinctly perform these string manipulation operations.

This answer will explore some of these string functions, suggest faster alternatives, and go into a timings comparison at the end.


.str.replace

Specify the substring/pattern to match, and the substring to replace it with.

pd.__version__
# '0.24.1'

df    
    time result
1  09:00   +52A
2  10:00   +62B
3  11:00   +44a
4  12:00   +30b
5  13:00  -110a

df['result'] = df['result'].str.replace(r'D', '')
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

If you need the result converted to an integer, you can use Series.astype,

df['result'] = df['result'].str.replace(r'D', '').astype(int)

df.dtypes
time      object
result     int64
dtype: object

If you don’t want to modify df in-place, use DataFrame.assign:

df2 = df.assign(result=df['result'].str.replace(r'D', ''))
df
# Unchanged

.str.extract

Useful for extracting the substring(s) you want to keep.

df['result'] = df['result'].str.extract(r'(d+)', expand=False)
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

With extract, it is necessary to specify at least one capture group. expand=False will return a Series with the captured items from the first capture group.


.str.split and .str.get

Splitting works assuming all your strings follow this consistent structure.

# df['result'] = df['result'].str.split(r'D').str[1]
df['result'] = df['result'].str.split(r'D').str.get(1)
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

Do not recommend if you are looking for a general solution.


If you are satisfied with the succinct and readable str
accessor-based solutions above, you can stop here. However, if you are
interested in faster, more performant alternatives, keep reading.


Optimizing: List Comprehensions

In some circumstances, list comprehensions should be favoured over pandas string functions. The reason is because string functions are inherently hard to vectorize (in the true sense of the word), so most string and regex functions are only wrappers around loops with more overhead.

My write-up, Are for-loops in pandas really bad? When should I care?, goes into greater detail.

The str.replace option can be re-written using re.sub

import re

# Pre-compile your regex pattern for more performance.
p = re.compile(r'D')
df['result'] = [p.sub('', x) for x in df['result']]
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

The str.extract example can be re-written using a list comprehension with re.search,

p = re.compile(r'd+')
df['result'] = [p.search(x)[0] for x in df['result']]
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

If NaNs or no-matches are a possibility, you will need to re-write the above to include some error checking. I do this using a function.

def try_extract(pattern, string):
    try:
        m = pattern.search(string)
        return m.group(0)
    except (TypeError, ValueError, AttributeError):
        return np.nan

p = re.compile(r'd+')
df['result'] = [try_extract(p, x) for x in df['result']]
df

    time result
1  09:00     52
2  10:00     62
3  11:00     44
4  12:00     30
5  13:00    110

We can also re-write @eumiro’s and @MonkeyButter’s answers using list comprehensions:

df['result'] = [x.lstrip('+-').rstrip('aAbBcC') for x in df['result']]

And,

df['result'] = [x[1:-1] for x in df['result']]

Same rules for handling NaNs, etc, apply.


Performance Comparison

enter image description here

Graphs generated using perfplot. Full code listing, for your reference. The relevant functions are listed below.

Some of these comparisons are unfair because they take advantage of the structure of OP’s data, but take from it what you will. One thing to note is that every list comprehension function is either faster or comparable than its equivalent pandas variant.

Functions

def eumiro(df):
    return df.assign(
        result=df['result'].map(lambda x: x.lstrip('+-').rstrip('aAbBcC')))

def coder375(df):
    return df.assign(
        result=df['result'].replace(r'D', r'', regex=True))

def monkeybutter(df):
    return df.assign(result=df['result'].map(lambda x: x[1:-1]))

def wes(df):
    return df.assign(result=df['result'].str.lstrip('+-').str.rstrip('aAbBcC'))

def cs1(df):
    return df.assign(result=df['result'].str.replace(r'D', ''))

def cs2_ted(df):
    # `str.extract` based solution, similar to @Ted Petrou's. so timing together.
    return df.assign(result=df['result'].str.extract(r'(d+)', expand=False))

def cs1_listcomp(df):
    return df.assign(result=[p1.sub('', x) for x in df['result']])

def cs2_listcomp(df):
    return df.assign(result=[p2.search(x)[0] for x in df['result']])

def cs_eumiro_listcomp(df):
    return df.assign(
        result=[x.lstrip('+-').rstrip('aAbBcC') for x in df['result']])

def cs_mb_listcomp(df):
    return df.assign(result=[x[1:-1] for x in df['result']])
Answered By: cs95

Try this using regular expression:

import re
data['result'] = data['result'].map(lambda x: re.sub('[-+A-Za-z]',x)
Answered By: Mr. Prophet

Suppose your DF is having those extra character in between numbers as well.The last entry.

  result   time
0   +52A  09:00
1   +62B  10:00
2   +44a  11:00
3   +30b  12:00
4  -110a  13:00
5   3+b0  14:00

You can try str.replace to remove characters not only from start and end but also from in between.

DF['result'] = DF['result'].str.replace('+|a|b|-|A|B', '')

Output:

  result   time
0     52  09:00
1     62  10:00
2     44  11:00
3     30  12:00
4    110  13:00
5     30  14:00
Answered By: Rishi Bansal

using "str.replace" is more fast than lambda and map when your data dimention is large:

your_data["result"]=your_data["result"].str.replace("+","")
your_data["result"]=your_data["result"].str.replace("-","")
Answered By: Ali karimi
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