[jure1]

Hi guys

I'm new in Python so i have an question.
My code work but i know that it's very very bad way to do it.I don't know what is best way to skip big number in for loop

My code:

Code

first = ["a", "b", "c", "d", "e", "f", "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","!","-","_","0","1","2","3","4","5","6","7","8","9","."]

    keywords = map(''.join, product(first, repeat=x))
    for key in keywords:
        asd += 1
        if (asd > 100000000):

I'm doing this to skip for loop 100000000 times...It's bad,i know..But what is faster way to skip it? I need to skip big number,because i don't know other way..

Thank you

[.kr.]

What is it for? What is it for?

[jure1]

I want to generate all possible character.But i want to skip some(because i'm generate it before..)

[Vitani]

This is probably what you need:

PHP Code


<?php 
first = ["a", "b", "c", "d", "e", "f", "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","!","-","_","0","1","2","3","4","5","6","7","8","9","."]

keywords = map(''.join, product(first, repeat=x))
for key in keywords:
    asd += 1
    if asd <= len(first): 


This basically says that if asd is smaller or equal to the length of first (40), it'll do whatever you put below the if statement line.

[QGK]

This is probably what you need:

PHP Code


<?php 
first = ["a", "b", "c", "d", "e", "f", "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","!","-","_","0","1","2","3","4","5","6","7","8","9","."]

keywords = map(''.join, product(first, repeat=x))
for key in keywords:
    asd += 1
    if asd <= len(first): 


This basically says that if asd is smaller or equal to the length of first (40), it'll do whatever you put below the if statement line.

I suppose so - the issue from OP is not really clear to me as to what he actually wants to do.

If all he has to do is check if the asd (basically an index now) is less than the length of the "first" list, it's unneccesary since the for loop already iterates over the list

[ChInfoOrg]

Hello!
There are differents possibilities to do Something like you want. you can use for exemple the "string" module

with something like this (as basic).

>>> import string
>>> string.ascii_uppercase
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> string.printable
'0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~
>>>

[michalmonday]

Code

my_list = ["first", "second", "third", "fourth"]
for item in my_list:
   print item

Outputs:
first
second
third
fourth

Code

my_list = ["first", "second", "third", "fourth"]
for item in my_list[2:]:
    print item

Outputs:
third
fourth

So apply the same to "keywords" list by saving the last checked keyword number (i) and then using it inside keywords[last_i_number:]. Save/keep track of last "asd" like:

Code

for key in keywords:
        asd += 1
        print("Last asd = " + str(asd))

And then reuse it like:

Code

for key in keywords[last_asd:]:

Or just save it to file and read it instead of changing the code manually each time, but that's the way in general

[vorticalbox]

This is probably what you need:

PHP Code


<?php 
first = ["a", "b", "c", "d", "e", "f", "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","!","-","_","0","1","2","3","4","5","6","7","8","9","."]

keywords = map(''.join, product(first, repeat=x))
for key in keywords:
    asd += 1
    if asd <= len(first): 


This basically says that if asd is smaller or equal to the length of first (40), it'll do whatever you put below the if statement line.

wouldn't this store them all in keywords and then loop over them? should it no be something like this?

PHP Code


<?php 
first = ["a", "b", "c", "d", "e", "f", "g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","!","-","_","0","1","2","3","4","5","6","7","8","9","."]
for key in map(''.join, product(first, repeat=x)):
    asd += 1
    if asd <= len(first):