How to split at specified delimiter

[SriRajesh]

Hi,
I have below DataFrame:
input:

VA004L200E2T99
VG002K500E4T99
VG002K500E4R45
VN009K30E6T00
VA007K100E2T40

I want to split at "E" and extract the numerical part to the right & left side of "E"
Desired output:

200E2
500E4
500E4
30E6
100E2

I tried the bleo code, but it is removing the specified character.

import re
out=re.sub(r'E', '', input)
print(out)

[metulburr]

you can split at the E, however you would have to insert the E again at some point

>>> 'VA004L200E2T99'.split('E')
['VA004L200', '2T99']

If the letter is always K or L, you could split the first again via that letter to parse out the 200; and the same for T and R. Or you could use regex, which i am horrible at.

[SriRajesh]

But I need to extract my output as:
200E2

[Gribouillis]

Or you could use regex, which i am horrible at.

Regex is not difficult to learn and it is very powerful for some tasks

>>> import re
>>> regex = re.compile(
... r"\d*" # zero or more digits
... r"E"   # the letter E
... r"\d*" # again zero or more digits
... )
>>> match = regex.search('VA004L200E2T99')
>>> match.group()
'200E2'

[SriRajesh]

I tried as below:

import re
regex = re.compile(r"\d*" r"\d*")
match = regex.search('VA004L200E2T99')
aa=match.group()
print(aa)

but it does not print give any output.

[snippsat]

but it does not print give any output.

He do it from interactive shell.
Like this without:

import re

regex = re.compile(r"\d*E\d*")
match = regex.search('VN009K30E6T00')
aa = match.group()
print(aa)

I have below DataFrame:

When using pandas,you should not take it out of DataFrame and use regex.
pandas has build in regex(str) to deal with DataFrame.
Example:

import pandas as pd

df = pd.DataFrame(
    [['VA004L200E2T99',77, 999],
     ['VG002K500E4T99',55, 100],
     ['VN009K30E6T00',33, 9]],
     columns=['val','foo', 'bar']
)

df['val'] = df['val'].str.extract(r'(\d*E\d*)', expand=False)
print(df)

Output:
     val  foo  bar
0  200E2   77  999
1  500E4   55  100
2   30E6   33    9