This is what I have:
>>> import random
>>> chars = [1, 6, 7, 8, 5.6, 3]
>>> for k in range(1, len(chars)+1):
... print random.choice(chars)
... chars[random.choice(chars)] = ''
...
But when I run it,
5.6
1
5.6
8
>>>
I don't want it to print a random amount of each one, I want it to print all of the content once, in a random order. And why is it printing spaces?
Try this:
import random
chars = [1, 6, 7, 8, 5.6, 3]
r = chars [:] #make a copy in order to leave chars untouched
random.shuffle (r) #shuffles r in place
print (r)
import random
chars = ['s', 'p', 'a', 'm']
random.shuffle(chars)
for char in chars:
print char
This will randomize the list in-place though.
list = [1, 6, 7]
random.shuffle(list)
for k in range(1,len(list)+1):
print list[k]
Here is a working version of your code.
import random
chars = [1, 6, 7, 8, 5.6, 3]
for k in range(1, len(chars)+1):
thechar = random.choice(chars)
place = chars.index(thechar)
print thechar
chars[place:place+1]=''
When you do the random.choice twice in print random.choice(chars) and in chars[random.choice(chars)] = '', it is picking another random choice from chars. Instead, set the random.choice to something so that you can call it later. Even if you had done that, when you set chars[random.choice(chars)] = '', it just sets that spot to '', it doesn't delete it. So if your random.choice was 5.6, the chars list would just become [1, 6, 7, 8, '', 3], not [1, 6, 7, 8, 3]. To do that, you have to save the place of the char, and do chars[place:place+1]. And finally, due to the for syntax, you have to do the len(chars)+1, so that it only goes to len(chars).
See the following link: Shuffle and re shuffle list in python?
from random import shuffle
x = [[i] for i in range(10)]
shuffle(x)
print x
