I have a list below:
list = [' ab: 1', ' cd: 0', ' ef: 2 gh: 3', ' ik: 4']
From above list, I need something like below:
newlist = [' ab: 1', ' cd: 0', ' ef: 2', ' gh: 3', ' ik: 4']
So "list" contains 4 elements and then I need "newlist" to contain 5 elements.
Is there any way split() can be used here?
I would approach this by using re.findall in the following way:
import re
li = [' ab: 1', ' cd: 0', ' ef: 2 gh: 3', ' ik: 4']
newlist = []
for ele in li:
match = re.findall(r"\w+\s?:\s?\d+", ele)
newlist += [m for m in match]
print(newlist)
Output:
['ab: 1', 'cd: 0', 'ef: 2', 'gh: 3', 'ik: 4']
Checking method for the best.
list_name = [' ab: 1', ' cd: 0', ' ef: 2 gh: 3', ' ik: 4']
newlist = []
for value in list_name:
if len(value) > 6:
newlist.append(value[0:6])
newlist.append(value[6:])
else:
newlist.append(value)
print(newlist)
as you can see, python have a great syntax called slicing (value[0:6])
if you learned about range(start,stop,step) you can use this as well list_name[start:stop:step]
the empty value will automaticaly set into the default [0,-1,1]
have fun :)
Try this:
newlist = []
for i in a:
if len(i.split()) != 1:
for key, val in zip(i.split()[::2], i.split()[1::2]):
newlist.append(key + val)
list!