Elementwise multiplication of characters in a list by numbers from a given shorter range while ignoring spaces and digits and joining result together

I want to multiply every letter of a word based on the starting and ending range given. My current code is as follows, but it doesn't work as intended:

word = ['h', 'e', 'l', 'p', 'm', 'e']
start = 3
end = 5
for b in range(start, end + 1):
    for a in word:
        print(a * b, end='')

My output is:

hhheeelllpppmmmeeehhhheeeellllppppmmmmeeeehhhhheeeeelllllpppppmmmmmeeeee

But I want to receive:

hhheeeelllllpppmmmmeeeee

The flow of this program should be like this:

helpme
345345

In addition, how could I ignore spaces and digits in the list? For example, my input would be:

['h', 'e', 'l', 'p', ' ', 'm', 'e']
start = 3
end = 5

And my output:

hhheeeelllllppp mmmmeeeee

so the multiplications should be:

help me
3453145

You could do something like the following, using zip and itertools.cycle:

import itertools

word = ['h', 'e', 'l', 'p', 'm', 'e']
start = 3
end = 5

for a, b in zip(itertools.cycle(range(start, end + 1)), word):
    print(b*a, end = '')

Output:

hhheeeelllllpppmmmmeeeee

Alternatively, you could do this using a modulo operator and enumerate:

word = ['h', 'e', 'l', 'p', 'm', 'e']
start = 3
end = 5

lst = list(range(start, end + 1))

for i, a in enumerate(word):
    print(a*lst[i%len(lst)], end='')

To ignore certain characters:

import itertools

word = ['h', 'e', 'l', 'p', ' ', '1', 'm', 'e']
start = 3
end = 5

characters_to_ignore=' 0123456789'

gen = itertools.cycle(range(start, end + 1))

lst = [(x, next(gen)) if x not in characters_to_ignore else (x, 1) for x in word]

for a, b in lst:
    print(a*b, end='')

Output:

hhheeeelllllppp 1mmmmeeeee

Using list comprehension and join:

word = "helpme"
start = 3
end = 5

r = list(range(start, end + 1))

print(''.join([l * r[i % len(r)] for i, l in enumerate(word)]))

Output

hhheeeelllllpppmmmmeeeee

Already good answers have been given. Just want to add mine which is one liner (just for a change). You should ideally use a well readable and separated code but as I said, this solution is to just bring out 'pythonic variety' that exists :-).

word = ['h', 'e', 'l', 'p', 'm', 'e']
print("".join(list(map(lambda a: a[0]*a[1], zip(word,list(range(3,6))*(len(word)//3))))))

Output is as below

hhheeeelllllpppmmmmeeeee

Lines added later

Edited after seeing a request of word list with space embedded in it

@Faded I can't think of a 'single line' solution without destroying sanity checks beyond repair. Everything has limits and I have hit the limit for producing 'one line' solution for the problem you have described. It could only be me and somebody might have very easy solution using single line python expression. We should wait for that special torch-bearer.

In the meanwhile, it is very easy to achieve the same with a multiline program. One out of many possible solution is as below.

word = ['h', 'e', 'l', 'p', '','m', 'e']
FinalList = []
MultiplierCount = 3
for i in word:
    if len(i) != 0:
        FinalList.append(i*MultiplierCount)
        MultiplierCount += 1
        if MultiplierCount > 5:
            MultiplierCount = 3
    else:
        FinalList.append(' ')

FinalOutput = "".join(FinalList)
print(FinalOutput)

It produces the following output.

hhheeeelllllppp mmmmeeeee

You can make use of enumerate and modulus operators here:

word = ['h', 'e', 'l', 'p', 'm', 'e']
start = 3
end = 5
for i,a in enumerate(word):
    print(a * (i%(end-start+1)+start), end='')

What's happening in your original code is that you're iterating through the whole word 3 times, once for each number in the range 3-6 exclusive. In this code, we use the index of the character mod 3 (end-start+1), and then add 3. So for any given number, it will give us a result of 3, 4, or 5.

EDIT:

I can't think of any logical way for you to use a nested for-loop to solve this without getting really complicated for no reason. You'll either end up iterating over range(3,6) for each letter of your word, or, like you're already doing, iterating through each letter of your word once for each number in range(3,6).