I have the below code
import re
age = []
txt = ('9', "10y", "4y",'unknown')
for t in txt:
if t.isdigit() is True:
age.append(re.search(r'\d+',t).group(0))
else:
age.append('unknown')
print(age)
and I get: ['9', 'unknown', 'unknown', 'unknown']
So the 9 I get, but I also need to get the 10 in the second position, the 4 in the third and unknown for the last.
Can anyone point me in the right direction?
Thank you for your help!
We can make use of the fact that re.search returns None when not finding any digit:
txt = ('9', "10y", "4y",'unknown')
age = []
for t in txt:
num = re.search('\d+', t)
if num:
age.append(num.group(0))
else:
age.append('unknown')
['9', '10', '4', 'unknown']
Since you tagged pandas, if you have a column, use str.extract:
pd.Series(txt).str.extract('(\d+)')
0 9
1 10
2 4
3 NaN
dtype: object
import re
age = []
txt = ('9', "10y22", "4y", 'unknown')
for t in txt:
res = re.findall('[0-9]+', t)
if res:
age.append(res[0])
else:
age.append("unknown")
import re
age = []
txt = ('9', "10y", "4y",'unknown')
for t in txt:
if len(t) > 1 and not t.isdigit():
t = t.replace(t[-1], '')
if t.isdigit() is True:
age.append(re.search(r'\d+',t).group(0))
else:
age.append('unknown')
print(age)
Check this out. So the len function checks if the string is bigger than one, and then if the last letter of the string is not a digit, then the string's last letter is being replaced with an empty space. And then it follows the rest of your algorithm. You can modify it more to fit your requirements, since you didn't specify that much.
