只获取最后的数字（正则表达式）

Question

I'd like to get only the numbers (integers) at the end of the phrases below:

VISTA AES TIETE E UNT N2 600 
VISTA IT AUUNIBANCO PN N1 1.400
OPCAO DE VENDA 04/21 COGNP450ON 4,50COGNE 100.000

I mean: 600, 1400, 100000. I'll add each one of them to a database later.

I tried to use regex: (?<=\s)(\d*\s*)|(\d*.\d*)$

But it didn't work properly. Any ideas?

PS: We use dots, not commas to represent a thousand: 1.000, instead of 1,000.

Answer 1

Actually for your use case, I don't think you'd even need regex

You can just split the strings and take the last one, and replace dot by empty string

If it's dataframe (since you have tagged Pandas ),

> df['colName'].str.split().str[-1].str.replace('.', '')
0       600
1      1400
2    100000
Name: colName, dtype: object

If it's list of strings

> list(map(lambda x: x.replace('.', ''),map(lambda x: x.split()[-1], data)))
['600', '1400', '100000']

Answer 2

l = ["VISTA AES TIETE E UNT N2 600",
"VISTA IT AUUNIBANCO PN N1 1.400",
"OPCAO DE VENDA 04/21 COGNP450ON 4,50COGNE 100.000"]

If the data is in the form of dataframe.

df=DataFrame({
    'col':l
})
df.col.str.extract('(\d*\.*\d*)?$').astype(str).replace('\.','', regex=True)

Output

0   600
1   1400
2   100000

Answer 3

In the pattern that you tried, this part (?<=\s)(\d*\s*) matches optional digits, followed by optional whitespace chars while there must be a whitespace char directly to the left.

That will also get all the positions in the string where there is a whitspace char to the left, as the digits and the whitespace char in the match are optional.

In this part (\d*\.\d*)$ the digits are optional, so it could also match just a dot at the end of the string.

---

If there has to be a whitespace char before the number at the end, you can use:

(?<=\s)\d{1,3}(?:\.\d{3})*$

The pattern matches:

• (?<=\s) Positive lookbehind, assert a whitspace char to the left from the current position
• \d{1,3} Match 1-3 digits
• (?:\.\d{3})* Optionally repeat a dot and 3 digits
• $ End of string

See a regex demo .

If the number can also be by itself, you could assert a whitespace boundary to the left (?<!\S)

(?<!\S)\d{1,3}(?:\.\d{3})*$

See another regex demo .

For example, using str.extract and wrapping the pattern in a capture group:

import pandas as pd

strings = [
    "VISTA AES TIETE E UNT N2 600",
    "VISTA IT AUUNIBANCO PN N1 1.400",
    "OPCAO DE VENDA 04/21 COGNP450ON 4,50COGNE 100.000"
]

df = pd.DataFrame(strings, columns=["colName"])
df['lastNumbers'] = df['colName'].str.extract(r"(?<=\s)(\d{1,3}(?:\.\d{3})*)$")

print(df)

Output

                                             colName lastNumbers
0                       VISTA AES TIETE E UNT N2 600         600
1                    VISTA IT AUUNIBANCO PN N1 1.400       1.400
2  OPCAO DE VENDA 04/21 COGNP450ON 4,50COGNE 100.000     100.000