how to replace non-numeric chars using regex

Question

I am wondering how to use regex remove any non-numeric chars while only selecting non-empty and spaces (a single value may contain one or multiple spaces) values for a series in a more efficient way,

df['numeric_no'] = df['id'].apply(lambda x: re.sub("[^0-9]", "", x))
df = df[(df['numeric_no'] != '') & (df['numeric_no'] != ' ')]

some sample data for the df

numeric_no
B-27000
44-11-E
LAND-11-4
17772A
88LL9A
321LP-3
UNIT 9 CAM -00-12
WWcard_055_34QE
EE119.45
aaa
b  b

the result will look like

Thank you, so all rows are selected? No one removed from sample? — jezrael
– jezrael, Commented May 23, 2018 at 11:33

jezrael · Accepted Answer · 2018-05-23 12:12:29Z

1

I believe need str.findall with boolean indexing:

s = df['numeric_no'].str.findall("(\d*\.\d+|\d+)").str.join('')

s = s[s.astype(bool)]
print (s)

0     27000
1      4411
2       114
3     17772
4       889
5      3213
6     90012
7     05534
8    119.45
Name: numeric_no, dtype: object

edited May 23, 2018 at 12:12

answered May 23, 2018 at 11:16

jezrael

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1 Comment

Bharath M Shetty Over a year ago

I think this is what OP needs but still needs some clarification

revo · Accepted Answer · 2018-05-23 11:58:27Z

1

You could match and capture numbers and match any thing else:

(\d+(?:\.\d+)?)|.

Live demo

Then replace match with $1 (a back-reference to first capturing group)

Python code:

re.sub(r"(\d+(?:\.\d+)?)|.", "$1", x)

answered May 23, 2018 at 11:58

revo

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Comments

Scott Boston · Accepted Answer · 2018-05-23 12:02:36Z

1

I think can try:

df.numeric_no.str.extractall('(\d+?[\.\d+])').astype(str).sum(level=0)

Output:

answered May 23, 2018 at 12:02

Scott Boston

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1 Comment

revo Over a year ago

Are you meaning \d+ in a character class? or each one separately? \d, +?

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