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Pandas replace function - insert pad 0 to 3 numbers

in Pandas (df), a column with following strings. looking to pad 0 when number within string are <100
Freq
XXX100KHz
XYC200KHz
YYY80KHz
YYY50KHz
to:
Freq
XXX100KHz
XYC200KHz
YYY080KHz
YYY050KHz
following function doesn't work, as \1 then 0 won't work as \10 doesn't exist.
df.replace({'Freq':'^([A-Za-z]+)(\d\d[A-Za-z]*)$'},{'Freq':r'\1**0**\2'},regex=True, inplace=True)

Try:
df["Freq"] = df["Freq"].str.replace(
r"(?<=\D)\d{1,2}(?=KHz)",
lambda g: "{:0>3}".format(g.group()),
regex=True,
)
print(df)
Prints:
Freq
0 XXX100KHz
1 XYC200KHz
2 YYY080KHz
3 YYY050KHz

Trying to format my values in pandas Dataframe which r showing like 1,008.59-, 1008.59 (one space is there aftr +ve value) to -1008.59 and 1008.59

Amtindoccurre
1008.59 (right space is there after this value)
1008.59-
Need to format those values in proper format like 1008.59(without any right side space)and -1008.59 in Pandas Data frame.
Can any one tell the way how to do.
Thanks
DS

You can use .str.replace() with regex to parse the float numbers with trailing sign and space and put the sign and space at the front. Then use .str.strip() to remove any space remaining:
data = {'val': ['1008.59 ', '1008.59-', '57,039.54 ', '4,232.49 ', '4,191.59-', '1,257,039.54 ', '2,257,039.54-']}
df = pd.DataFrame(data)
df['val'] = df['val'].str.replace(r'(\d+(?:,\d+)*(?:\.\d+)?)(\s|-)', r'\2\1', regex=True).str.strip()
Result:
print(df)
val
0 1008.59
1 -1008.59
2 57,039.54
3 4,232.49
4 -4,191.59
5 1,257,039.54
6 -2,257,039.54
If you want to remove the thousand separators , also, you can use:
df['val'].str.replace(r'(\d+(?:,\d+)*(?:\.\d+)?)(\s|-)', r'\2\1', regex=True).str.strip().str.replace(',', '', regex=True)
Result:
print(df)
val
0 1008.59
1 -1008.59
2 57039.54
3 4232.49
4 -4191.59
5 1257039.54
6 -2257039.54
If you want to further convert the numbers from string to numeric values (float type), you can use:
df['val'] = df['val'].str.replace(r'(\d+(?:,\d+)*(?:\.\d+)?)(\s|-)', r'\2\1', regex=True).str.strip().str.replace(',', '', regex=True).astype(float)

How to split a string without given delimeter in Panda

dfcolumn = [PUEF2CarmenXFc034DpEd, PUEF2BalulanFc034CamH, CARF1BalulanFc013Baca, ...]
My output should be:
dfnewcolumn1 = [PUEF2, PUEF2 , CARF1]
dfnewcolumn2 = [CarmenXFc034DpEd, BalulanFc034CamH, BalulanFc013Baca]

Assuming your split criteria is by fixed number of characters (e.g. 5 here), you can use:
df['dfnewcolumn1'] = df['dfcolumn'].str[:5]
df['dfnewcolumn2'] = df['dfcolumn'].str[5:]
Result:
dfcolumn dfnewcolumn1 dfnewcolumn2
0 PUEF2CarmenXFc034DpEd PUEF2 CarmenXFc034DpEd
1 PUEF2BalulanFc034CamH PUEF2 BalulanFc034CamH
2 CARF1BalulanFc013Baca CARF1 BalulanFc013Baca
If your split criteria is by the first digit in the string, you can use:
df[['dfnewcolumn1', 'dfnewcolumnX']] = df['dfcolumn'].str.split(r'(?<=\d)\D', n=1, expand=True)
df[['dfnewcolumnX', 'dfnewcolumn2']] = df['dfcolumn'].str.split(r'\D*\d', n=1, expand=True)
df = df.drop(columns='dfnewcolumnX')
Using the following modified original data with more test cases:
dfcolumn
0 PUEF2CarmenXFc034DpEd
1 PUEF2BalulanFc034CamH
2 CARF1BalulanFc013Baca
3 CAF1BalulanFc013Baca
4 PUEFA2BalulanFc034CamH
Run code:
df[['dfnewcolumn1', 'dfnewcolumnX']] = df['dfcolumn'].str.split(r'(?<=\d)\D', n=1, expand=True)
df[['dfnewcolumnX', 'dfnewcolumn2']] = df['dfcolumn'].str.split(r'\D*\d', n=1, expand=True)
df = df.drop(columns='dfnewcolumnX')
Result:
dfcolumn dfnewcolumn1 dfnewcolumn2
0 PUEF2CarmenXFc034DpEd PUEF2 CarmenXFc034DpEd
1 PUEF2BalulanFc034CamH PUEF2 BalulanFc034CamH
2 CARF1BalulanFc013Baca CARF1 BalulanFc013Baca
3 CAF1BalulanFc013Baca CAF1 BalulanFc013Baca
4 PUEFA2BalulanFc034CamH PUEFA2 BalulanFc034CamH

Assuming your prefix consists of a sequence of alphabets followed by a sequence of digits, which both have variable length. Then a regex split function can be constructed and applied on each cell.
Solution
import pandas as pd
import re
# data
df = pd.DataFrame()
df["dfcolumn"] = ["PUEF2CarmenXFc034DpEd", "PUEF2BalulanFc034CamH", "CARF1BalulanFc013Baca"]
def f_split(s: str):
"""Split two part by regex"""
# alphabet(s) followed by digit(s)
o = re.match(r"^([A-Za-z]+\d+)(.*)$", s)
# may add exception handling here if there is no match
return o.group(1), o.group(2)
df[["dfnewcolumn1", "dfnewcolumn2"]] = df["dfcolumn"].apply(f_split).to_list()
Note the .to_list() to convert tuples into lists, which is required for the new column assignment to work.
Result
print(df)
dfcolumn dfnewcolumn1 dfnewcolumn2
0 PUEF2CarmenXFc034DpEd PUEF2 CarmenXFc034DpEd
1 PUEF2BalulanFc034CamH PUEF2 BalulanFc034CamH
2 CARF1BalulanFc013Baca CARF1 BalulanFc013Baca

Hoe about this compact solution:
import pandas as pd
df = pd.DataFrame({"original": ["PUEF2CarmenXFc034DpEd", "PUEF2BalulanFc034CamH", "CARF1BalulanFc013Baca"]})
df2 = pd.DataFrame(df.original.str.split(r"(\d)", n=1).to_list(), columns=["part1", "separator", "part2"])
df2.part1 = df2.part1 + df2.separator.astype(str)
df2
part1 separator part2
0 PUEF2 2 CarmenXFc034DpEd
1 PUEF2 2 BalulanFc034CamH
2 CARF1 1 BalulanFc013Baca
I use:
Series.str.split with a regex pattern and a kwarg to specify that it should only split on the first match.
in th regex pattern, I use a group (the round braces in (\d)) to capture the separating character
to_list() to output the split as a list of lists
DataFrame constructor to build a new DataFrame from that list
string concat of two columns

I want to remove digits and parenthesis from a data column in dataframe

can anyone make me understand this piece of code.
def remove_digit(data):
newData = ''.join([i for i in data if not i.isdigit()])
i = newData.find('(')
if i>-1: newData = newData[:i]
return newData.strip()

Why don't you use regex. [0-9()] looks for matching characters between 0-9, ( and )
newData = re.sub('[0-9()]', '', data)

Give this df:
data
0 a43
1 b((
2 cr3r3
3 d
You can remove digits and parenthesis from the column in this way:
df['data'] = df['data'].str.replace('\d|\(|\)','')
Output:
data
0 a
1 b
2 crr
3 d

Pandas DataFrame get substrings from column

I have a column named "KL" with for example:
sem_0405M4209F2057_1.000
sem_A_0103M5836F4798_1.000
Now I want to extract the four digits after "M" and the four digits after "F". But with df["KL"].str.extract I can't get it to work.
Locations of M and F vary, thus just using the slice [9:13] won't work for the complete column.

If you want to use str.extract, here's how:
>>> df['KL'].str.extract(r'M(?P<M>[0-9]{4})F(?P<F>[0-9]{4})')
M F
0 4209 2057
1 5836 4798
Here, M(?P<M>[0-9]{4}) matches the character 'M' and then captures 4 digits following it (the [0-9]{4} part). This is put in the column M (specified with ?P<M> inside the capturing group). The same thing is done for F.

You could use split to achieve this, probably a better way exists:
In [147]:
s = pd.Series(['sem_0405M4209F2057_1.000','sem_A_0103M5836F4798_1.000'])
s
Out[147]:
0 sem_0405M4209F2057_1.000
1 sem_A_0103M5836F4798_1.000
dtype: object
In [153]:
m = s.str.split('M').str[1].str.split('F').str[0][:4]
f = s.str.split('M').str[1].str.split('F').str[1].str[:4]
print(m)
print(f)
0 4209
1 5836
dtype: object
0 2057
1 4798
dtype: object

You can also use regex:
import re
def get_data(x):
data = re.search( r'M(\d{4})F(\d{4})', x)
if data:
m = data.group(1)
f = data.group(2)
return m, f
df = pd.DataFrame(data={'a': ['sem_0405M4209F2057_1.000', 'sem_0405M4239F2027_1.000']})
df['data'] = df['a'].apply(lambda x: get_data(x))
>>
a data
0 sem_0405M4209F2057_1.000 (4209, 2057)
1 sem_0405M4239F2027_1.000 (4239, 2027)