1

How can i use a for loop inside a while loop? Here is my code:

def avoids(word,forbidden):
    for fl in forbidden:
        for letter in word:
            if letter == fl:
                return False
    return True

fin= open('words.txt')
u=97
v=97
w=97
x=97
y=97
minim=100
while u <= 122:
    while v <= 122:
        while w <= 122:
            while x <= 122:
                while y <= 122:
                    count=0
                    for line in fin:
                        word = line.strip()
                        if avoids(word,chr(u)+chr(v)+chr(w)+chr(x)+chr(y)):
                            #print(word)
                            count+=1
                            #print((100/113809)*count)
                    if (100/113809)*count<minim:
                        print(count)
                        minim=(100/113809)*count
                        print(minim,chr(u)+chr(v)+chr(w)+chr(x)+chr(y))
                    y+=1
                y=97
                x+=1
            x=97
            w+=1
        w=97
        v+=1
    v=97
    u+=1

It executes the for loop for just one time. I can put fin= open('words.txt') inside the latest while statement but then program gets really slow & almost unusable. What can i do?(not that i don't want to use lists & etc.)

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6
  • 3
    It's so slow when you put fin = open('words.txt') inside the last while loop because you are opening and reading the file 11,881,376 times... Are you sure that's how many loops you actually want?? Commented Jun 28, 2013 at 13:34
  • 6
    What exactly are you trying to do? Commented Jun 28, 2013 at 13:34
  • 1
    Please refactor this code because it breaks several good programming patterns. Code is unclear, contains semantic unknown constants defined, too much loop nesting, too big circular complexity, etc. Commented Jun 28, 2013 at 13:51
  • *jonhopkins Yes i know how many loops i want! *Tim I'm trying to create every possible combination of 5 English letters and send it to avoids function. This is a exercise(9.3) in the Thinkpython book.*This is just a temporary code and i'm a beginner so please excuse me *sgnsajgon. Commented Jun 28, 2013 at 14:13
  • 1
    Please please PLEASE provide names for your magic numbers - it will help with understanding and maintaining code. For example, 97 and 122 should both be constants. Commented Jun 28, 2013 at 14:15

3 Answers 3

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5

The reason it is executing the for loop just one time is that you are exhausting the buffer you created for your "words.txt" file during the first iteration of the for loop.

If you want to go through the words in that file multiple times you need to reopen it each time (which, as you noted, creates a lot of overhead).

Alternatively, read that file into a list and then run the while/for-loop structure that list.

I.E.

fin= open('words.txt')
wordList = fin.readlines()
u=97
v=97
...
for line in wordList
...
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4 Comments

Thanks, but i don't really get it.why putting it to a list solves the problem of the for loop?*and my code still is slow. is there any way to speed it up?
When you are iterating through the fin itself it effectively 'destroys' the variable after going through it. Your for loop is just taking data out of fin until it reaches the end... but then there is no beginning to go back to when the for loop runs again, because you had already 'used up' all the data.
The basic task here is going to be time intensive. You're generating a lot of excess operations though, and I'm not sure this is going to give the answer you're expecting if you are just trying to generate every 5-character string possible.
Thank you...it seems that there is no other way to do it with the tools that i know.so i think i got the answer.
1

Your code would look a lot less indented like this:

from string import ascii_lowercase
from itertools import product

for u, v, w, x, y in product(ascii_lowercase, repeat=5):
    ...

I'm not sure what the avoids() function is supposed to do. It's unlikely to be useful in it's current form. Did you test it at all?

maybe your intent is something like this

def avoids(word, forbidden):
    for fl, letter in zip(forbidden, word):
        if letter == fl:
            return False
    return True

but it's hard to imagine how that would be useful. The logic still seems wrong

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2 Comments

These are all good and true points, but I'm not sure you answered his question.
Thank you,but same as the code that cdhagmanns has wrote, i don't know how it works... and i don't even know what zip does.but i have test it and it gives the result that i want. i don't really know what is wrong with it.but i may be wrong...
0

You can check words against a list much quicker than you could against a file as it skips both reading and writing overhead. This list can be form quickly using list comprehension.

import string
chars = string.uppercase
word_list = [''.join((a,b,c,d,e)) for a in chars for b in chars for c in chars
                                  for d in chars for e in chars]

'dkbke'.upper() in word_list
>>> True

You can come up with the rest as I not sure what all you want to do with it.

EDIT: As gnibbler just taught me, the above and be shorten using

from itertools import product
from string import uppercase as chars

words = [''.join((a,b,c,d,e)) for a, b, c, d, e in product(chars, repeat=5)]

'dkbke'.upper() in words
>>> True

NOTE: As for learning things with which you aren't familiar, try using __doc__ to learn or just play with it. Example:

product.__doc__
>>> product(*iterables) --> product object

Cartesian product of input iterables.  
Equivalent to nested for-loops.

For example, product(A, B) returns the same as:  
((x,y) for x in A for y in B).

The leftmost iterators are in the outermost for-loop, so the output tuples
cycle in a manner similar to an odometer (with the rightmost element changing
on every iteration).

To compute the product of an iterable with itself,
specify the number of repetitions with the optional repeat keyword argument.
For example,
product(A, repeat=4) means the same as product(A, A, A, A).

product('ab', range(3)) --> ('a',0) ('a',1) ('a',2) ('b',0) ('b',1) ('b',2)
product((0,1), (0,1), (0,1)) --> (0,0,0) (0,0,1) (0,1,0) (0,1,1) (1,0,0) ...

''.join.__doc__
>>> S.join(iterable) -> string

Return a string which is the concatenation of the strings in the
iterable.  The separator between elements is S.

''.join(['a','b','c'])
>>> 'abc'
'-'.join(['a','b','c'])
>>> 'a-b-c'
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3 Comments

Thank you, but i don't really know how your code works and since i should do it with what i know, i don't really want to learn them right now. i'm trying to make it work with basic tools...
While I agree that you should always only use code that you understand, I find that writing verbose code to do things that Python has built-in will never make you a better at Python. I would encourage you to play around with the code and learn what it does, even if you don't like if for this case.
OK, i will take a look into that. also thanks for the doc, it seems very useful.

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