Split list into two parts based on some delimiter in each list element in python
split_items = (i.split('/') for i in my_list)
my_list1, my_list2 = zip(*split_items)
This creates 2 tuples. If you really need lists, you can convert them by
my_list1, my_list2 = map(list, (my_list1, my_list2))
Splitting list into multiple lists - depending on elements index
You could split the strings in a list comprehension and use zip
:
list(zip(*[i.split(';') for i in combinedList]))
[('40% Football', '30% Basketball', '20% Baseball', '10% Rugby'),
('40% Football', '30% Basketball', '20% Base-Ball', '10% Le Rugby'),
('40% Fuball', '30% Basketball', '20% Baseball', '10% Rugby'),
('40% Futbol', '30% Baloncesto', '20% Béisbol', '10% Rugby'),
('40% Calcio', '30% Pallacanestro', '', '10% Rugby')]
Split a list in python with an element as the delimiter?
When you write,
new_list = [x.split('a')[-1] for x in l]
you are essentially performing,
result = []
for elem in l:
result.append(elem.split('a')[-1])
That is, you're splitting each string contained in l
on the letter 'a'
, and collecting the last element of each of the strings into the result.
Here's one possible implementation of the mechanic you're looking for:
def extract_parts(my_list, delim):
# Locate the indices of all instances of ``delim`` in ``my_list``
indices = [i for i, x in enumerate(my_list) if x == delim]
# Collect each end-exclusive sublist bounded by each pair indices
sublists = []
for i in range(len(indices)-1):
part = my_list[indices[i]+1:indices[i+1]]
sublists.append(part)
return sublists
Using this function, we have
>>> l = ['a', 'b', 'c', 'c', 'b', 'a', 'b', 'c', 'b', 'a']
>>> extract_parts(l, 'a')
[['b', 'c', 'c', 'b'], ['b', 'c', 'b']]
How to split elements in list?
If you have a small list you can try this:
lst = ['Name0, Location0', 'Phone number0', 'Name1, Location1', 'Phone number1']
s = ", "
joined = s.join(lst)
newList = joined.split(s)
print(newList)
Output:
['Name0', 'Location0', 'Phone number0', 'Name1', 'Location1', 'Phone number1']
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