Otdfbt

One way would be to do this:

s = 'Pythonisanprogramminglanguage'

arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']

expected = []
i = 0
for word in arr:
 expected.append(s[i:i+len(word)])
 i+= len(word)

print(expected)

Using a simple for loop this can be done as follows:

s = 'Pythonisanprogramminglanguage'

arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']

start_index = 0
expected = list()
for a in arr:
 expected.append(s[start_index:start_index+len(a)])
 start_index += len(a)

print(expected)

In the future, an alternative approach will be to use an assignment expression (new in Python 3.8):

s = 'Pythonisanprogramminglanguage' 
arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']

i = 0
expected = [s[i:(i := i+len(word))] for word in arr]

You can use itertools.accumulate to get the positions where you want to split the string:

>>> s = 'Pythonisanprogramminglanguage'
>>> arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']
>>> import itertools
>>> L = list(itertools.accumulate(map(len, arr)))
>>> L
[6, 8, 10, 21, 29]

Now if you zip the list with itself, you get the intervals:

>>> list(zip([0]+L, L))
[(0, 6), (6, 8), (8, 10), (10, 21), (21, 29)]

And you just have to use the intervals to split the string:

>>> [s[i:j] for i,j in zip([0]+L, L)]
['Python', 'is', 'an', 'programming', 'language']

Create a simple loop and use the length of the words as your index:

s = 'Pythonisanprogramminglanguage' 
arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']

ctr = 0
words = []
for x in arr:
 words.append(s[ctr:len(x) + ctr])
 ctr += len(x)

print(words)

# ['Python', 'is', 'an', 'programming', 'language']

The itertools module has a function named accumulate() (added in Py 3.2) to help make this relatively easy:

from itertools import accumulate # added in Py 3.2


s = 'Pythonisanprogramminglanguage'
arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']

cuts = tuple(accumulate(len(item) for item in arr))
words = [s[i:j] for i, j in zip((0,)+cuts, cuts)]
print(words) # -> ['Python', 'is', 'an', 'programming', 'language']

Here is another approach :

import numpy as np
ar = [0]+list(map(len, arr))
ar = list(np.cumsum(ar))
output_ = [s[i:ar[ar.index(i)+1]] for i in ar[:-1]]

Output :

['Python', 'is', 'an', 'programming', 'language']

One more way

a,l = 0,[]
for i in map(len,arr):
 l.append(s[a:a+i])
 a+=i
print (l)
#['Python', 'is', 'an', 'programming', 'language']

Props to the answer using iter. The accumulate answers are my favorite. Here is another accumulate answer using map instead of a list comprehension

import itertools

s = 'Pythonisanprogramminglanguage'
arr = ['lkjhgf', 'zx', 'qw', 'ertyuiopakk', 'foacdhlc']
ticks = itertools.accumulate(map(len, arr[0:]))
words = list(map(lambda i, x: s[i:len(x) + i], (0,) + tuple(ticks), arr))

Output:

['Python', 'is', 'an', 'programming', 'language']

You could collect slices off the front of s.

output = []

for word in arr:
 i = len(word)
 chunk, s = s[:i], s[i:]
 output.append(chunk)

print(output) # -> ['Python', 'is', 'an', 'programming', 'language']

Yet another approach would be to create a regex pattern describing the desired length of words. You can replace every character by . (=any character) and surround the words with ():

arr = ['lkjhgf', 'zx', 'q', 'ertyuiopakk', 'foacdhlc']

import re

pattern = '(' + ')('.join(re.sub('.', '.', word) for word in arr) + ')'
#=> '(......)(..)(.)(...........)(........)'

If the pattern matches, you get the desired words in groups directly:

s = 'Pythonisaprogramminglanguage'
re.match(pattern, s).groups()
#=> ('Python', 'is', 'a', 'programming', 'language')