Split list by cost and keep separators
I have a list called list_of_strings
that looks like this:
['a', 'b', 'c', 'a', 'd', 'c', 'e']
I want to split this list by value (in this case c
). I also want to keep c
the resulting split.
So the expected output is:
[['a', 'b', 'c'], ['a', 'd', 'c'], ['e']]]
Any easy way to do this?
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You can use more_itertoools
to accomplish this simply and clearly:
from more_itertools import split_after
lst = ["a", "b", "c", "a", "d", "c", "e"]
list(split_after(lst, lambda x: x == "c"))
# [['a', 'b', 'c'], ['a', 'd', 'c'], ['e']]
Another example: here we are splitting words by simply changing the predicate :
lst = ["ant", "bat", "cat", "asp", "dog", "carp", "eel"]
list(split_after(lst, lambda x: x.startswith("c")))
# [['ant', 'bat', 'cat'], ['asp', 'dog', 'carp'], ['eel']]
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def spliter(value, array):
res = []
while value in array:
index = array.index(value)
res.append(array[:index + 1])
array = array[index + 1:]
if array:
# Append last elements
res.append(array)
return res
a = ['a', 'b', 'c', 'a', 'd', 'c', 'e']
print(spliter('b',a))
# [['a', 'b'], ['c', 'a', 'd', 'c', 'e']]
print(spliter('c',a))
# [['a', 'b', 'c'], ['a', 'd', 'c'], ['e']]
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You can try something like the following:
list_of_strings = ['a', 'b', 'c', 'a', 'd', 'c', 'e']
output = [[]]
for x in list_of_strings:
output[-1].append(x)
if x == 'c':
output.append([])
Although it should be noted that this will add an empty list to your output if your last input element is 'c'
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How about this. It only needs to iterate over one input, and some of them are in a method index
that is executed as native code.
def splitkeep(v, c):
curr = 0
try:
nex = v.index(c)
while True:
yield v[curr: (nex + 1)]
curr = nex + 1
nex += v[curr:].index(c) + 1
except ValueError:
if v[curr:]: yield v[curr:]
print(list(splitkeep( ['a', 'b', 'c', 'a', 'd', 'c', 'e'], 'c')))
result
[['a', 'b', 'c'], ['a', 'd', 'c'], ['e']]
I was not sure if you wanted to keep an empty list at the end of the result if the final value was the value you split. I made the assumption that you won't, so I put the condition in a final value exception if it is empty.
This causes the input to []
only lead to []
, if possible, it can lead to [[]]
.
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How about this rather playful script:
a = ['a', 'b', 'c', 'a', 'd', 'c', 'e']
b = ''.join(a).split('c') # ['ab', 'ad', 'e']
c = [x + 'c' if i < len(b)-1 else x for i, x in enumerate(b)] # ['abc', 'adc', 'e']
d = [list(x) for x in c if x]
print(d) # [['a', 'b', 'c'], ['a', 'd', 'c'], ['e']]
It can also handle start and end values with "c"
a = ['c', 'a', 'b', 'c', 'a', 'd', 'c', 'e', 'c']
d -> [['c'], ['a', 'b', 'c'], ['a', 'd', 'c'], ['e', 'c']]
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list_of_strings = ['a', 'b', 'c', 'a', 'd', 'c', 'e']
value = 'c'
new_list = []
temp_list = []
for item in list_of_strings:
if item is value:
temp_list.append(item)
new_list.append(temp_list[:])
temp_list.clear()
else:
temp_list.append(item)
if (temp_list):
new_list.append(temp_list)
print(new_list)
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stuff = ['a', 'b', 'c', 'a', 'd', 'c', 'e']
You can find out the indices using 'c'
like this and add 1 because after that you will split it, not by its index:
indices = [i + 1 for i, x in enumerate(stuff) if x == 'c']
Then extract the snippets like this:
split_stuff = [stuff[i:j] for i, j in zip([0] + indices, indices + [None])]
zip
gives you a list of tuples, similar (indices[i], indices[i + 1])
, with the concatenated [0]
one allowing you to extract the first part and [None]
extract the last fragment ( stuff[i:]
)
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