python pandas replacing column values conditional on string patterns and using split() - regex

Question

long time lurker--I finally stuck to a project involving pandas and more than ever I need your help.
I have a dataframe like the following. Each row describe one retirement formula which may have more than one criteria (hence e1)
index e0 e1
1 62/10 NaN
2 age 55 NaN
3 67/10 age 70
I want to make a column age that describes the minimum age. I've defined patterns for how each criterion is described. For example,
pattern1=r'.*/.*'
pattern7=r'age.[0-9].*'
and I have pattern1-pattern7.
I used the following code to extract age portion of e0 to a new column age:
df['age']=df['e0'][(df['e0'].str.match(pattern1)==1)].apply(lambda x: str(x).split('/')[0])
which gives me
index e0 e1 age
1 62/10 NaN 62
2 age 55 NaN NaN
3 67/10 age 70 67
I want to address other formats such as "age 55" (to extract 55, in this case), but I'm not sure how to go about. If I do
df['age']=df['e0'][(df['e0'].str.match(pattern7)==1)].apply(lambda x: str(x).split(' ')[1])
then it's clearly wrong because I'd overwrite what's already in age and I get
index e0 e1 age
1 62/10 NaN NaN
2 age 55 NaN 55
3 67/10 age 70 NaN
I've tried other variations as far as the syntax would allow me but to no avail.
I'm a Stata user and in Stata, I'd be using replace command conditional on regexm. I'm trying to learn Python and it's been a difficult journey! I'd appreciate any help on this.
I have another (hopefully) quick question in addition: I've used the following two lines to get rid of white space in both e0 and e1.
option['e0']=option['e0'].str.strip()
option['e1']=option['e1'].str.strip()
Is there a way to address them both in one line?
Thanks a lot in advance.

Answer 1

This is a response to your second question (you should stick to one question per post).
df.loc[:, ['e0', 'e1']].apply(lambda x: x.str.strip())
I'm not sure why you are calling the DataFrame 'option' when it was previously referred to as 'df', so I stuck with the latter.

Answer 2

Interesting problem, here I pass a function that removes the NaN values and then calls sum which will concatenate the rows of data.
We can then call the vectorised str method findall with regex \d+ which returns all numbers as a list.
We then apply another function to this that converts the str numbers to ints, puts these in a list and returns the smallest value:
In [37]:
def func(x):
return x.dropna().sum()
​
def lowest(x):
return min(list(map(int,x)))
​
df['min'] = df[['e0','e1']].apply(lambda x: func(x), axis=1).str.findall(r'\d+').apply(lowest)
df
Out[37]:
index e0 e1 min
0 1 62/10 NaN 10
1 2 age 55 NaN 55
2 3 67/10 age 70 10
Breaking the above down so you can see what is happening:
In [38]:
df[['e0','e1']].apply(lambda x: func(x), axis=1)
Out[38]:
0 62/10
1 age 55
2 67/10age 70
dtype: object
In [39]:
df[['e0','e1']].apply(lambda x: func(x), axis=1).str.findall(r'\d+')
Out[39]:
0 [62, 10]
1 [55]
2 [67, 10, 70]
dtype: object