How do i get list of numbers?

Question

a=("9,223;789:52 88:65;2,78:55")
b= ""
for char in a:
if not char.isnumeric():
b=b+char
print(b)
for char in a:
if char not in b:
value="".join(char).split()
else:
" "
for val in value:
print(int(val))

i wrote this code.,but i didn't get my desire output.
Answer should be=[9,223,789,52,88,65,2,78,55]
what should i do?

Answer 1

Gouri Shinde ,

i have done a try with regex split function. it allows to define multiple separators:

import re
a=("9,223;789:52 88:65;2,78:55")
print([int(num) for num in re.split(';|,|:|\s',a)])

result is: [9, 223, 789, 52, 88, 65, 2, 78, 55]

>>> [edited]:
▪︎If the list contains several consecutive identical separators, the above mentioned code will create an error. to fix this, the pattern string has to be modified.

import re
a=("9,223;789::::52 88:65;2,,,,78:55")
print([int(num) for num in re.split(';+|,+|:+|\s+',a)])

this will create a correct result. it is done by adding a '+', to each separator. that means that it searches for "1 or more" of the respective preceding characters.

Answer 2

import re
a=("9,223;789:52 88:65;2,78:55")
b=re.sub("[;: ]",",",a)
c=b.split(",")
d=list(map(lambda x:int(x),c))
print(d)

Answer 3

How about this solution?
https://code.sololearn.com/cDkrNy2tg77E/?ref=app

Answer 4

lines 7 to 13, judging by what you wrote I think you originally meant this:

"".join([char if char not in b else " " for char in a]).split()

print([int(val) for val in value])

Answer 5

a=("9,223;789:52 88:65;2,78:55")
b = value = " "
be=[]
for char in a:
if not char.isnumeric():
b=b+char
if char not in b:
value += char
else:
value += " "
#or:
"""
for char in a:
if not char.isnumeric():
b += char
value += char if char not in b else" "
"""
for val in value.split():
be += [int(val)]
print(be)