String separation in space areas-CodePudding

we have a string with spaces and we have max string length, i.e.

str1 = 'we have a string with spaces and we have max string length'
b=8

And we need to divide string in rows with max b length in row, but we can divide only in space areas, so in our example it would be like:

x='
we have
a string
with
spaces
and we
have max
string
length'

and it should work for any b. I have tried to make a list from a string and check can two elements of list be less than b(but the can be 3 or 4 etc, so dont know what to do in this case). !And very important that by the task description b is actually named as 'len' , so i believe this is made to prevent using this function...

str1 = 'we have a string with spaces and we have max string length'
def WordSearch(len:int, a:str->int:
    k=[]
    t=str1.split(sep=' ')
    z=str()
    lst1 = ['wp'   str(i) for i in range(len(t))]
    for j in lst1:
        for i in range(len(t)):
            if len(j) len(t[i])<lenn:
                'wp0' t[0] ' '

    print(lst1)
WordSearch(len=8,a=str1)

Is it possible to fix code? or i am going completely wrong way....

CodePudding user response：

No need to reinvent the wheel:

import textwrap
output = textwrap.wrap('we have a string with spaces and we have max string length', 8)
output == ['we have', 'a string', 'with', 'spaces', 'and we', 'have max', 'string', 'length']

You can combine the items with "\n".join(output), print them, or any other way of your choosing.

CodePudding user response：

It looks like some algorithm exercise, anyway it just need to take care of corner cases.

str1 = 'we have a string with spaces and we have max string length'
def WordSearch(len:int, a:str):
    k=[]
    first_str = ''
    first_len = 0
    second_str = ''
    second_len = 0
    is_first = True
    for c in a:
        if c == ' ':
            if is_first:
                is_first = False
                continue
            if first_len   second_len   1 > len:
                k.append(first_str)
                first_str = second_str
                first_len = second_len
                second_str = ''
                second_len = 0
            else:
                first_str = first_str   ' '   second_str
                first_len  = second_len   1
                second_str = ''
                second_len = 0
        else:
            if is_first:
                first_str  = c
                first_len  = 1
            else:
                second_str  = c
                second_len  = 1

    if second_len != 0:
        if first_len != 0:
            if (first_len   second_len   1 > len):
                k.append(first_str)
                k.append(second_str)
            else:
                k.append(first_str   ' '   second_str)
        else:
            k.append(second_str)
    elif first_len != 0:
        k.append(first_str)
    result = '\r\n'.join(k)
    print(result)
WordSearch(len=8,a=str1)

CodePudding user response：

If you really want to write it yourself (rather than using textwrap) then:

def reformat(s, b):
    tokens = s.split()
    if len(tokens) > 0 and max(map(len, tokens)) <= b:
        result = [[tokens[0]]]
        for token in tokens[1:]:
            e = result[-1]
            if len(token)   sum(map(len, e))   len(e) <= b:
                e.append(token)
            else:
                result.append([token])
        return [' '.join(e) for e in result]
s = 'we have a string with spaces and we have max string length'
print(*reformat(s, 8), sep='\n')

Output:

we have
a string
with
spaces
and we
have max
string
length