I want to replace a string if I don't have a specific delimiter/string in a column. If the delimiter already exists in that row then I don't want to touch that row. I have around 3.5million records.
Below is the sample set. I want to replace is with :
One:1
Two:2
Three is 3
Four is IV:4
Output should be like this
One:1
Two:2
Three:3
Four is IV:4
Try it without loop and a one-liner using loc
df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4", "Five is V"], columns=["myValues"])
df.loc[~df['myValues'].str.contains(':'), 'myValues'] = df.loc[~df['myValues'].str.contains(':'), 'myValues'].str.replace('is', ':')
print(df)
myValues
0 One:1
1 Two:2
2 Three : 3
3 Four is IV:4
4 Five : V
Option 1
Inplace with update
df.update(
df.myValues.loc[
lambda x: ~x.str.contains(':')
].str.replace('\s+is\s+', ':'))
myValues
0 One:1
1 Two:2
2 Three:3
3 Four is IV:4
Option 2
Inline and using map
f = lambda x: x if ':' in x else x.replace(' is ', ':')
df.assign(myValues=list(map(f, v)))
myValues
0 One:1
1 Two:2
2 Three:3
3 Four is IV:4
Other than .contains(), you can also use simple string operations:
df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4"], columns=["myValues"])
target = [":" not in e for e in df.myValues]
df.myValues[target] = df.myValues[target].str.replace(" is ",":")
Result:
myValues
0 One:1
1 Two:2
2 Three:3
3 Four is IV:4
First, filter out all strings containing :. Then, replace " is " with ":" for all rows left. (In your example, the space around "is" is also deleted. Thus, I replace " is " with ":".)
df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4"], columns=["myValues"])
for idx, v in df[~df.myValues.str.contains(":")].iterrows():
df.loc[idx].myValues = df.iloc[idx].myValues.replace(" is ", ":")
References