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
Related
How can i extract the values within the quotes signs into two separate columns with python. The dataframe is given below:
df = pd.DataFrame(["'FRH02';'29290'", "'FRH01';'29300'", "'FRT02';'29310'", "'FRH03';'29340'",
"'FRH05';'29350'", "'FRG02';'29360'"], columns = ['postcode'])
df
postcode
0 'FRH02';'29290'
1 'FRH01';'29300'
2 'FRT02';'29310'
3 'FRH03';'29340'
4 'FRH05';'29350'
5 'FRG02';'29360'
i would like to get an output like the one below:
postcode1 postcode2
FRH02 29290
FRH01 29300
FRT02 29310
FRH03 29340
FRH05 29350
FRG02 29360
i have tried several str.extract codes but havent been able to figure this out. Thanks in advance.
Finishing Quang Hoang's solution that he left in the comments:
import pandas as pd
df = pd.DataFrame(["'FRH02';'29290'",
"'FRH01';'29300'",
"'FRT02';'29310'",
"'FRH03';'29340'",
"'FRH05';'29350'",
"'FRG02';'29360'"],
columns = ['postcode'])
# Remove the quotes and split the strings, which results in a Series made up of 2-element lists
postcodes = df['postcode'].str.replace("'", "").str.split(';')
# Unpack the transposed postcodes into 2 new columns
df['postcode1'], df['postcode2'] = zip(*postcodes)
# Delete the original column
del df['postcode']
print(df)
Output:
postcode1 postcode2
0 FRH02 29290
1 FRH01 29300
2 FRT02 29310
3 FRH03 29340
4 FRH05 29350
5 FRG02 29360
You can use Series.str.split:
p1 = []
p2 = []
for row in df['postcode'].str.split(';'):
p1.append(row[0])
p2.append(row[1])
df2 = pd.DataFrame()
df2["postcode1"] = p1
df2["postcode2"] = p2
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
I have a dataframe column with names:
df = pd.DataFrame({'Names': ['ROS-053', 'ROS-54', 'ROS-51', 'ROS-051B', 'ROS-051A', 'ROS-52']})
df.replace(to_replace=r'[a-zA-Z]{3}-\d{2}$', value='new', regex=True)
The format needs to be three letters followed by - then three numbers. So ROS-51 should be replaced with ROS-051.. And ROS-051B should be ROS-051. I have tried numerous things but can't seem to figure it out.
Any help would be highly appreciated:)
You can do:
df['Names'] = df.Names.replace('^([a-zA-Z]{3})-0?(\d{2})(.*)$', r'\1-0\2', regex=True)
Output:
Names
0 ROS-053
1 ROS-054
2 ROS-051
3 ROS-051
4 ROS-051
5 ROS-052
Here is one option using regex replacement with a callback:
repl = lambda m: m.group(1) + ('00' + m.group(2))[-3:] + m.group(3)
df.str.replace(r'^([A-Z]{3}-)(\d+)(.*)$', repl)
Note this answer is flexible and will left pad with zeroes either a single or double digit only to three digits.
Here's another way to do it:
df = pd.DataFrame({'Names': ['ROS-053', 'ROS-54', 'ROS-51', 'ROS-051B', 'ROS-051A', 'ROS-52']})
df['Names'] = df['Names'].str.replace(r'[A-Z]$', '')
df['Names'] = df['Names'].str.split('-').str[0] + '-' + df['Names'].str.split('-').str[1].apply(lambda x: x.zfill(3))
print(df)
Output:
Names
0 ROS-053
1 ROS-054
2 ROS-051
3 ROS-051
4 ROS-051
5 ROS-052
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
I have currently run the following script which uses Fuzzylogic to replace some common words from the list. Dataframe df1 contains my default list of possible values. Dataframe df2 is the main dataframe where transformations/changes are undertaken after referring to Dataframe df1. The code is as follows:
df1 = pd.DataFrame(['one','two','three','four','five','tsst'])
df2 = pd.DataFrame({'not_shifted':[np.nan,'one','too','three','fours','five','six',np.nan,'test']})
# Drop nan value
df2=pd.DataFrame(df2['not_shifted'].fillna(value=''))
df2['not_shifted'] = df2['not_shifted'].map(lambda x: difflib.get_close_matches(x, df1[0]))
The problem is the output is a dataframe which contains square brackets. To make matters worse, none of the texts within df2['not_shifted'] are viewable/ recallable:
Out[421]:
not_shifted
0 []
1 [one]
2 [two]
3 [three]
4 [four]
5 [five]
6 []
7 []
8 [tsst]
Please help.
df2.not_shifted.apply(lambda x: x[0] if len(x) != 0 else "") or simply df2.not_shifted.str[0] as solved by #Psidom
def replace_all(eg):
rep = {"[":"",
"]":"",
"u":"",
"}":"",
"'":"",
'"':"",
"frozenset":""}
for i,j in rep.items():
eg = eg.replace(i,j)
return eg
for each in df.columns:
df[each] = df[each].apply(lambda x : replace_all(str(x)))