python pandas melt on combined columns - pandas

Question

I have a dataframe like this. i have regular fields till "state" then i will have trailers (3 columns tr1* represents 1 tailer) i want to convert those trailers to rows. I tried melt function but i am able to use only 1 trailer column. kindly look at below example you can understand
Name number city state tr1num tr1acct tr1ct tr2num tr2acct tr2ct tr3num tr3acct tr3ct
DJ 10 Edison nj 1001 20345 Dew 1002 20346 Newca. 1003. 20347. pen
ND 20 Newark DE 2001 1985 flor 2002 1986 rodge
I am expecting the output like this.
Name number city state trnum tracct trct
DJ 10 Edison nj 1001 20345 Dew
DJ 10 Edison nj 1002 20346 Newca
DJ 10 Edison nj 1003 20347 pen
ND 20 Newark DE 2001 1985 flor
ND 20 Newark DE 2002 1986 rodge

Answer 1

You need to look at using pd.wide_to_long. However, you will need to do some column renaming first.
df = df.set_index(['Name','number','city','state'])
df.columns = df.columns.str.replace('(\D+)(\d+)(\D+)',r'\1\3_\2')
df = df.reset_index()
pd.wide_to_long(df, ['trnum','trct','tracct'],
['Name','number','city','state'], 'Code',sep='_',suffix='\d+')\
.reset_index()\
.drop('Code',axis=1)
Output:
Name number city state trnum trct tracct
0 DJ 10 Edison nj 1001.0 Dew 20345.0
1 DJ 10 Edison nj 1002.0 Newca. 20346.0
2 DJ 10 Edison nj 1003.0 pen 20347.0
3 ND 20 Newark DE 2001.0 flor 1985.0
4 ND 20 Newark DE 2002.0 rodge 1986.0
5 ND 20 Newark DE NaN NaN NaN

Answer 2

you could achieve this by renaming your columns and bit and applying the pandas wide_to_long method. Below is the code which produces your desired output.
df = pd.DataFrame({"Name":["DJ", "ND"], "number":[10,20], "city":["Edison", "Newark"], "state":["nj","DE"],
"trnum_1":[1001,2001], "tracct_1":[20345,1985], "trct_1":["Dew", "flor"], "trnum_2":[1002,2002],
"trct_2":["Newca", "rodge"], "trnum_3":[1003,None], "tracct_3":[20347,None], "trct_3":["pen", None]})
pd.wide_to_long(df, stubnames=['trnum', 'tracct', 'trct'], i='Name', j='dropme', sep='_').reset_index().drop('dropme', axis=1)\
.sort_values('trnum')
outputs
Name state city number trnum tracct trct
0 DJ nj Edison 10 1001.0 20345.0 Dew
1 DJ nj Edison 10 1002.0 NaN Newca
2 DJ nj Edison 10 1003.0 20347.0 pen
3 ND DE Newark 20 2001.0 1985.0 flor
4 ND DE Newark 20 2002.0 NaN rodge
5 ND DE Newark 20 NaN NaN None

Answer 3

Another option:
df = pd.DataFrame({'col1': [1,2,3], 'col2':[3,4,5], 'col3':[5,6,7], 'tr1':[0,9,8], 'tr2':[0,9,8]})
The df:
col1 col2 col3 tr1 tr2
0 1 3 5 0 0
1 2 4 6 9 9
2 3 5 7 8 8
subsetting to create 2 df's:
tr1_df = df[['col1', 'col2', 'col3', 'tr1']].rename(index=str, columns={"tr1":"tr"})
tr2_df = df[['col1', 'col2', 'col3', 'tr2']].rename(index=str, columns={"tr2":"tr"})
res = pd.concat([tr1_df, tr2_df])
result:
col1 col2 col3 tr
0 1 3 5 0
1 2 4 6 9
2 3 5 7 8
0 1 3 5 0
1 2 4 6 9
2 3 5 7 8

Answer 4

One option is the pivot_longer function from pyjanitor, using the .value placeholder:
# pip install git+https://github.com/pyjanitor-devs/pyjanitor.git
import janitor
import pandas as pd
(df
.pivot_longer(
index=slice("Name", "state"),
names_to=(".value", ".value"),
names_pattern=r"(.+)\d(.+)",
sort_by_appearance=True)
.dropna()
)
Name number city state trnum tracct trct
0 DJ 10 Edison nj 1001.0 20345.0 Dew
1 DJ 10 Edison nj 1002.0 20346.0 Newca.
2 DJ 10 Edison nj 1003.0 20347.0 pen
3 ND 20 Newark DE 2001.0 1985.0 flor
4 ND 20 Newark DE 2002.0 1986.0 rodge
The .value keeps the part of the column associated with it as header, and since we have multiple .value, they are combined into a single word. The .value is determined by the groups in the names_pattern, which is a regular expression.
Note that currently the multiple .value option is available in dev.