Extract a text out of a column using pattern in python - python-3.x

Question

I'm trying to extract a text out of a column so I can move to another column using a pattern in python but I miss some results at the same time I need to keep the unextracted strings as they are>
My code is:
import pandas as pd
df = pd.DataFrame({
'col': ['item1 (30-10)', 'item2 (200-100)', 'item3 (100 FS)', 'item4 (100+)', 'item1 (1000-2000)' ]
})
pattern = r'(\d+(\,[0-9]+)?\-\d+(\,[a-zA-Z])?\d+)'
df['result'] = df['col'].str.extract(pattern)[0]
print(df)
My output is:
col result
0 item1 (30-10) 30-10
1 item2 (200-100) 200-100
2 item3 (100 FS) NaN
3 item4 (100+) NaN
4 item1 (1000-2000) 1000-2000
My output should be:
col result newcolumn
0 item1 (30-10)
1 item2 (200-100)
2 item3 (100 FS)
3 item4 (100+)
4 item1 (1000-2000)

Answer 1

You can use this:
df['newcolumn'] = df.col.str.extract(r'(\(.+\))')
df['result'] = df['col'].str.extract(r'(\w+)')
Output:
col newcolumn result
0 item1 (30-10) (30-10) item1
1 item2 (200-100) (200-100) item2
2 item3 (100 FS) (100 FS) item3
3 item4 (100+) (100+) item4
4 item1 (1000-2000) (1000-2000) item1
Explanation:
The first expression gets the content within parenthesis (including the parenthesis themselves). The second gets the first word.

Answer 2

You can also do this with .str.split in a single line:
df[['result', 'newcolumn']] = df['col'].str.split(' ', 1, expand=True)
Output:
col result newcolumn
0 item1 (30-10) item1 (30-10)
1 item2 (200-100) item2 (200-100)
2 item3 (100 FS) item3 (100 FS)
3 item4 (100+) item4 (100+)
4 item1 (1000-2000) item1 (1000-2000)
You must use expand=True if your strings have a non-uniform number of splits (see also How to split a dataframe string column into two columns?).
EDIT: If you want to 'drop' the old column, you can also overwrite it and rename it:
df[['col', 'newcolumn']] = df['col'].str.split(' ', 1, expand=True)
df = df.rename(columns={"col": "result"})
which exactly gives you the result you specified was intended:
result newcolumn
0 item1 (30-10)
1 item2 (200-100)
2 item3 (100 FS)
3 item4 (100+)
4 item1 (1000-2000)

Answer 3

You can extract the parts of interest by grouping them within one regular expression. The regex pattern now matches item\d as first group and anything inside the brackets with \(.*\) as the second one.
import pandas as pd
df = pd.DataFrame({
'col': ['item1 (30-10)', 'item2 (200-100)', 'item3 (100 FS)', 'item4 (100+)', 'item1 (1000-2000)' ]
})
pattern = "(item\d*)\s(\(.*\))"
df['items'] = df['col'].str.extract(pattern)[0]
df['result'] = df['col'].str.extract(pattern)[1]
print(df)
Output:
col items result
0 item1 (30-10) item1 (30-10)
1 item2 (200-100) item2 (200-100)
2 item3 (100 FS) item3 (100 FS)
3 item4 (100+) item4 (100+)
4 item1 (1000-2000) item1 (1000-2000)