We Keep Coding

Trying to make a new column with existing column that have 'int' and 'string'

data = pd.read_csv(r'RE_absentee_one.csv')
data['New_addy'] = str(data['Prop-House Number']) + data['Prop-Street Name'] + data['Prop-Mode'] + str(data['Prop-Apt Unit Number'])
df = pd.DataFrame(data, columns = ['Name','New_addy'])
So this is the code
As you can see Prop-House Number and Prop-Apt Number are both int, and the rest are strings, I am trying to combine all these so that the full address is under one column labeled 'New addy'

Follow the string assignment with each variable using map as mentioned below:
data = pd.read_csv(r'RE_absentee_one.csv')
data['New_addy'] = data['Prop-House Number'].map(str) + data['Prop-Street Name'].map(str) + data['Prop-Mode'].map(str) + data['Prop-Apt Unit Number'].map(str)
#select the desired columns for further work
data = data[['Name','New_addy']]

One way is using list comprehension:
data['New_addy'] = [str(n) + street + mode + str(apt_n) for n,street,mode,apt_n in zip(
data['Prop-House Number'],data['Prop-Street Name'],data['Prop-Mode'],data['Prop-Apt Unit Number'])]

Concatenate on specific condition python

EDITED
I want to write an If loop with conditions on cooncatenating strings.
i.e. If cell A1 contains a specific format of text, then only do you concatenate, else leave as is.
example:
If bill number looks like: CM2/0000/, then concatenate this string with the date column (month - year), else leave the bill number as it is.
Sample Data

You can create function which does what you need and use df.apply() to execute it on all rows.
I use example data from #Boomer answer.
EDIT: you didn't show what you really have in dataframe and it seems you have datetime in bill_date but I used strings. I had to convert strings to datetime to show how to work with this. And now it needs .strftime('%m-%y') or sometimes .dt.strftime('%m-%y') instead of .str[3:].str.replace('/','-'). Because pandas uses different formats to display dateitm for different countries so I couldn't use str(x) for this because it gives me 2019-09-15 00:00:00 instead of yours 15/09/19
import pandas as pd
df = pd.DataFrame({
'bill_number': ['CM2/0000/', 'CM2/0000', 'CM3/0000/', 'CM3/0000'],
'bill_date': ['15/09/19', '15/09/19', '15/09/19', '15/09/19']
})
df['bill_date'] = pd.to_datetime(df['bill_date'])
def convert(row):
if row['bill_number'].endswith('/'):
#return row['bill_number'] + row['bill_date'].str[3:].replace('/','-')
return row['bill_number'] + row['bill_date'].strftime('%m-%y')
else:
return row['bill_number']
df['bill_number'] = df.apply(convert, axis=1)
print(df)
Result:
bill_number bill_date
0 CM2/0000/09-19 15/09/19
1 CM2/0000 15/09/19
2 CM3/0000/09-19 15/09/19
3 CM3/0000 15/09/19
Second idea is to create mask
mask = df['bill_number'].str.endswith('/')
and later use it for all values
#df.loc[mask,'bill_number'] = df[mask]['bill_number'] + df[mask]['bill_date'].str[3:].str.replace('/','-')
df.loc[mask,'bill_number'] = df[mask]['bill_number'] + df[mask]['bill_date'].dt.strftime('%m-%y')
or
#df.loc[mask,'bill_number'] = df.loc[mask,'bill_number'] + df.loc[mask,'bill_date'].str[3:].str.replace('/','-')
df.loc[mask,'bill_number'] = df.loc[mask,'bill_number'] + df.loc[mask,'bill_date'].dt.strftime('%m-%y')
Left side needs .loc[mask,'bill_number'] instead of `[mask]['bill_number'] to correctly assing values - but right side doesn't need it.
import pandas as pd
df = pd.DataFrame({
'bill_number': ['CM2/0000/', 'CM2/0000', 'CM3/0000/', 'CM3/0000'],
'bill_date': ['15/09/19', '15/09/19', '15/09/19', '15/09/19']
})
df['bill_date'] = pd.to_datetime(df['bill_date'])
mask = df['bill_number'].str.endswith('/')
#df.loc[mask,'bill_number'] = df[mask]['bill_number'] + df[mask]['bill_date'].str[3:].str.replace('/','-')
# or
#df.loc[mask,'bill_number'] = df.loc[mask,'bill_number'] + df.loc[mask,'bill_date'].str[3:].str.replace('/','-')
df.loc[mask,'bill_number'] = df[mask]['bill_number'] + df[mask]['bill_date'].dt.strftime('%m-%y')
#or
#df.loc[mask,'bill_number'] = df.loc[mask,'bill_number'] + df.loc[mask,'bill_date'].dt.strftime('%m-%y')
print(df)
Third idea is to use numpy.where()
import pandas as pd
import numpy as np
df = pd.DataFrame({
'bill_number': ['CM2/0000/', 'CM2/0000', 'CM3/0000/', 'CM3/0000'],
'bill_date': ['15/09/19', '15/09/19', '15/09/19', '15/09/19']
})
df['bill_date'] = pd.to_datetime(df['bill_date'])
df['bill_number'] = np.where(
df['bill_number'].str.endswith('/'),
#df['bill_number'] + df['bill_date'].str[3:].str.replace('/','-'),
df['bill_number'] + df['bill_date'].dt.strftime('%m-%y'),
df['bill_number'])
print(df)

Maybe this will work for you. It would be nice to have a data sample like #Mike67 was stating. But based on your information this is what I came up with. Bulky, but it works. I'm sure someone else will have a fancier version.
import pandas as pd
from pandas import DataFrame, Series
dat = {'num': ['CM2/0000/','CM2/0000', 'CM3/0000/', 'CM3/0000',],
'date': ['15/09/19','15/09/19','15/09/19','15/09/19']}
df = pd.DataFrame(dat)
df['date'] = df['date'].map(lambda x: str(x)[3:])
df['date'] = df['date'].str.replace('/','-')
for cols in df.columns:
df.loc[df['num'].str.endswith('/'), cols] = df['num'] + df['date']
print(df)
Results:
num date
0 CM2/0000/09-19 09-19
1 CM2/0000 09-19
2 CM3/0000/09-19 09-19
3 CM3/0000 09-19

Formatting a specific row of integers to the ssn style

I want to format a specific column of integers to ssn format (xxx-xx-xxxx). I saw that openpyxl has builtin styles. I have been using pandas and wasn't sure if it could do this specific format.
I did see this -
df.iloc[:,:].str.replace(',', '')
but I want to replace the ',' with '-'.
import pandas as pd
df = pd.read_excel('C:/Python/Python37/Files/Original.xls')
df.drop(['StartDate', 'EndDate','EmployeeID'], axis = 1, inplace=True)
df.rename(columns={'CheckNumber': 'W/E Date', 'CheckBranch': 'Branch','DeductionAmount':'Amount'},inplace=True)
df = df[['Branch','Deduction','CheckDate','W/E Date','SSN','LastName','FirstName','Amount','Agency','CaseNumber']]
ssn = (df['SSN'] # the integer column
.astype(str) # cast integers to string
.str.zfill(8) # zero-padding
.pipe(lambda s: s.str[:2] + '-' + s.str[2:4] + '-' + s.str[4:]))
writer = pd.ExcelWriter('C:/Python/Python37/Files/Deductions Report.xlsx')
df.to_excel(writer,'Sheet1')
writer.save()

Your question is a bit confusing, see if this helps:
If you have a column of integers and you want to create a new one made up of strings in SSN (Social Security Number) format. You can try something like:
df['SSN'] = (df['SSN'] # the "integer" column
.astype(int) # the integer column
.astype(str) # cast integers to string
.str.zfill(9) # zero-padding
.pipe(lambda s: s.str[:3] + '-' + s.str[3:5] + '-' + s.str[5:]))

Setup
Social Security numbers are nine-digit numbers using the form: AAA-GG-SSSS
s = pd.Series([111223333, 222334444])
0 111223333
1 222334444
dtype: int64
Option 1
Using zip and numpy.unravel_index:
pd.Series([
'{}-{}-{}'.format(*el)
for el in zip(*np.unravel_index(s, (1000,100,10000)))
])
Option 2
Using f-strings:
pd.Series([f'{i[:3]}-{i[3:5]}-{i[5:]}' for i in s.astype(str)])
Both produce:
0 111-22-3333
1 222-33-4444
dtype: object

I prefer:
df["ssn"] = df["ssn"].astype(str)
df["ssn"] = df["ssn"].str.strip()
df["ssn"] = (
df.ssn.str.replace("(", "")
.str.replace(")", "")
.str.replace("-", "")
.str.replace(" ", "")
.apply(lambda x: f"{x[:3]}-{x[3:5]}-{x[5:]}")
)
This take into account rows that are partially formatted, fully formatted, or not formatted and correctly formats them all.
For Example:
data = [111111111,123456789,"222-11-3333","433-3131234"]
df = pd.DataFrame(data, columns=['ssn'])
Gives you:
Before
After the code you then get:
After

splitting string a multiple delimiters

Is there a way to split a string in a pandas dataframe column like
coordinates(gDNA)
chr10:g.89711916T>A
into tab separated fields
chr\start\ref\alt
chr10\t89711916\tT\tA
in pandas.
So far I've tried
df[['chr','others']] = df['coordinates(gDNA)'].str.split(':',expand=True)
and extracted the first part, but not sure what to do for the rest

Use:
df[['chr','start', 'alt']] = df['coordinates(gDNA)'].str.split(':g.|>',expand=True)
df[['start','ref']] = df['start'].str.extract('(\d+)(\D+)')
print (df)
coordinates(gDNA) chr start alt ref
0 chr10:g.89711916T>A chr10 89711916 A T

Try this:
df[['chr','start','ref','alt']] = df['coordinates(gDNA)'].str.extract('(\w+).*?(\d+)(\w+).*?(\w+)')

df = pd.DataFrame(
columns=['coordinates(gDNA)'],
data=[['chr10:g.89711916T>A']]
)
def parser(x):
ch, x = x.split(':g.')
start = int(x[:-3])
ref = x[-3]
alt = x[-1]
return dict(chr=ch, start=start, ref=ref, alt=alt)
pd.DataFrame([*map(parser, df['coordinates(gDNA)'])], df.index)
alt chr ref start
0 A chr10 T 89711916

Split and Join Series in Pandas

I have two series in the dataframe below. The first is a string which will appear in the second, which will be a url string. What I want to do is change the first series by concatenating on extra characters, and have that change applied onto the second string.
import pandas as pd
#import urlparse
d = {'OrigWord' : ['bunny', 'bear', 'bull'], 'WordinUrl' : ['http://www.animal.com/bunny/ear.html', 'http://www.animal.com/bear/ear.html', 'http://www.animal.com/bull/ear.html'] }
df = pd.DataFrame(d)
def trial(source_col, dest_col):
splitter = dest_col.str.split(str(source_col))
print type(splitter)
print splitter
res = 'angry_' + str(source_col).join(splitter)
return res
df['Final'] = df.applymap(trial(df.OrigWord, df.WordinUrl))
I'm trying to find the string from the source_col, then split on that string in the dest_col, then effect that change on the string in dest_col. Here I have it as a new series called Final but I would rather inplace. I think the main issue are the splitter variable, which isn't working and the application of the function.
Here's how result should look:
OrigWord WordinUrl
angry_bunny http://www.animal.com/angry_bunny/ear.html
angry_bear http://www.animal.com/angry_bear/ear.html
angry_bull http://www.animal.com/angry_bull/ear.html

apply isn't really designed to apply to multiple columns in the same row. What you can do is to change your function so that it takes in a series instead and then assigns source_col, dest_col to the appropriate value in the series. One way of doing it is as below:
def trial(x):
source_col = x["OrigWord"]
dest_col = x['WordinUrl' ]
splitter = str(dest_col).split(str(source_col))
res = splitter[0] + 'angry_' + source_col + splitter[1]
return res
df['Final'] = df.apply(trial,axis = 1 )

here is an alternative approach:
df['WordinUrl'] = (df.apply(lambda x: x.WordinUrl.replace(x.OrigWord,
'angry_' + x.OrigWord), axis=1))
In [25]: df
Out[25]:
OrigWord WordinUrl
0 bunny http://www.animal.com/angry_bunny/ear.html
1 bear http://www.animal.com/angry_bear/ear.html
2 bull http://www.animal.com/angry_bull/ear.html

Instead of using split, you can use the replace method to prepend the angry_ to the corresponding source:
def trial(row):
row.WordinUrl = row.WordinUrl.replace(row.OrigWord, "angry_" + row.OrigWord)
row.OrigWord = "angry_" + row.OrigWord
return row
df.apply(trial, axis = 1)
OrigWord WordinUrl
0 angry_bunny http://www.animal.com/angry_bunny/ear.html
1 angry_bear http://www.animal.com/angry_bear/ear.html
2 angry_bull http://www.animal.com/angry_bull/ear.html