In my python script, I have a list of strings like,
birth_year = ["my birth year is *","i born in *","i was born in *"]
I want to compare one input sentence with the above list and need a birth year as output.
The input sentence is like:
Example1: My birth year is 1994.
Example2: I born in 1995
The output will be:
Example1: 1994
Example2: 1995
I applied many approaches by using regex. But I didn't find a perfect solution for the same.
If you change birth_year to a list of regexes you could match more easily with your input string. Use a capturing group for the year.
Here's a function that does what you want:
def match_year(birth_year, input):
for s in birth_year:
m = re.search(s, input, re.IGNORECASE)
if m:
output = f'{input[:m.start(0)]}{m[1]}'
print(output)
break
Example:
birth_year = ["my birth year is (\d{4})","i born in (\d{4})","i was born in (\d{4})"]
match_year(birth_year, "Example1: My birth year is 1994.")
match_year(birth_year, "Example2: I born in 1995")
Output:
Example1: 1994
Example2: 1995
You need at least Python 3.6 for f-strings.
str1=My birth year is 1994.
str2=str1.replace('My birth year is ','')
You can try something like this and replace the unnecessary string with empty string.
For the code you shared, you can do something like :
for x in examples:
for y in birth_year:
if x.find(y)==1: #checking if the substring exists in example
x.replace(y,'') #if it exists we replace it with empty string
I think the above code might work
If you can guaranty those "strings like" always contain one 4 digits number, which is a year of birth, somewhere in there... i'd say just use regex to get whatever 4 digits in there surrounded by non-digits. Rather dumb, but hey, works with your data.
import re
examples = ["My birth year is 1993.", "I born in 1995", "я родился в 1976м году"]
for str in examples:
y = int(re.findall(r"^[^\d]*([\d]{4})[^\d]*$", str)[0])
print(y)
re.findall(r'(\d+)',val)[0]