Odtnhj

I have a DataFrame and I'd like to make only specific parts of strings to be made uppercase with an underscore afterwords.

| TYPE | NAME |
|-----------------------------|
| Contract Employee | John |
| Full Time Employee | Carol |
| Temporary Employee | Kyle |

I'd like the words "Contract" and "Temporary" made into uppercase like this with an underscore after and before the word:

| TYPE | NAME |
|-------------------------------|
| _CONTRACT_ Employee | John |
| Full Time Employee | Carol |
| _TEMPORARY_ Employee | Kyle |

I tried using str.upper() but that made the entire cell uppercase and I'm looking only for those certain words.

EDIT: I should mention sometimes the words are not capitalized if that matters. Often it will display as temporary employee instead of Temporary Employee.

Here is one option using re.sub:

def type_to_upper(match):
 return match.group(1).upper()

text = "Contract Employee"
output = re.sub(r'b(Contract|Temporary)b', type_to_upper, text)

EDIT:

This is the same approach applied within pandas, also addressing the latest edit regarding uncertain upper or lower case words to be replaced:

test dataframe:

 TYPE NAME
0 Contract Employee John
1 Full Time Employee Carol
2 Temporary Employee Kyle
3 contract employee John
4 Full Time employee Carol
5 temporary employee Kyle

solution:

def type_to_upper(match):
 return '__'.format(match.group(1).upper())

df.TYPE = df.TYPE.str.replace(r'b([Cc]ontract|[Tt]emporary)b', type_to_upper)

result:

df 
 TYPE NAME
0 _CONTRACT_ Employee John
1 Full Time Employee Carol
2 _TEMPORARY_ Employee Kyle
3 _CONTRACT_ employee John
4 Full Time employee Carol
5 _TEMPORARY_ employee Kyle

Note that this is only for addressing exactly these two cases which are defined in the OPs request. For complete case insensitivity it's even simpler:

df.TYPE = df.TYPE.str.replace(r'b(contract|temporary)b', type_to_upper, case=False)

Something that modifies the data-frame (without regex or anything):

l=['Contract','Temporary']
df['TYPE']=df['TYPE'].apply(lambda x: ' '.join(['_'+i.upper()+'_' if i in l else i for i in x.split()]))

join and split, being in a apply.

And then now:

print(df)

Is:

 TYPE NAME
0 _CONTRACT_ Employee John
1 Full Time Employee Carol
2 _TEMPORARY_ Employee Kyle

This is simples and easy way by using replace with dictionary format.

Please refer pandas Doc for Series.replace

df["TYPE"] = df["TYPE"].replace('Contract': '_CONTRACT_', 'Temporary': '_Temporary_', regex=True)

Just reproduced:

>>> df
 TYPE Name
0 Contract Employee John
1 Full Time Employee Carol
2 Temporary Employee Kyle

>>> df["TYPE"] = df["TYPE"].replace('Contract': '_CONTRACT_', 'Temporary': '_TEMPORARY_', regex=True)
>>> df
 TYPE Name
0 _CONTRACT_ Employee John
1 Full Time Employee Carol
2 _TEMPORARY_ Employee Kyle

U9 beat me to it, using lambda and split() on the input:

def match_and_upper(match):
 matches = ["Contract", "Temporary"]
 if match in matches:
 return match.upper()
 return match

input = "Contract Employee"
output = " ".join(map(lambda x: match_and_upper(x), input.split()))
# Result: CONTRACT Employee # 

Answering part of my own question here. Using @Tim Biegeleisen's regex he provided, I did a string replace on the column.

df["TYPE"] = df["TYPE"].str.replace(r'b(Contract)b', '_CONTRACT_')