Python Nested For Loop Counting Backwards

Question

This is probably a simple problem, but I am trying to created a nested loop that would count up from 0 to 9 in the outer loop, and in the inner loop, start from the value (or index. They are the same in this case) of the outer loop and count backwards.

Here's an example:

i= 0
k= 0

i= 1
k= 1
k= 0 

i= 2
k= 2
k= 1
k= 0

i= 3
k= 3
k= 2
k= 1
k= 0

I got this far:

x = range(0,10)
for i in x:
    print 'i = ',x[i]
    for k in x:
        print 'k = ', x[i::-1]

Obviously, the code above doesn't do what I want it to do. For one, the second for loop doesn't start from the value of i in the outer loop and counts backwards. For another, it doesn't print a new k = for every new value.

I assume here that k[i::-1] is really x[i::-1], isn't it? — Ricardo Cárdenes
– Ricardo Cárdenes, Commented Feb 4, 2015 at 23:46
Or you can do range( i, -1, -1 ) to get [ i, i-1, i-2, ...., 2, 1, 0] — sarveshseri
– sarveshseri, Commented Feb 4, 2015 at 23:51

Marcin · Accepted Answer · 2015-02-04 23:49:05Z

2

I think this should be like this:

x = range(0,10)
for i in x:
    print 'i = ',x[i]      
    for k in x[i::-1]:
        print 'k = ', k
    print("\n")

The result is:

i =  0
k =  0


i =  1
k =  1
k =  0


i =  2
k =  2
k =  1
k =  0


i =  3
k =  3
k =  2
k =  1
k =  0


i =  4
k =  4
k =  3
k =  2
k =  1
k =  0


i =  5
k =  5
k =  4
k =  3
k =  2
k =  1
k =  0


i =  6
k =  6
k =  5
k =  4
k =  3
k =  2
k =  1
k =  0


i =  7
k =  7
k =  6
k =  5
k =  4
k =  3
k =  2
k =  1
k =  0


i =  8
k =  8
k =  7
k =  6
k =  5
k =  4
k =  3
k =  2
k =  1
k =  0


i =  9
k =  9
k =  8
k =  7
k =  6
k =  5
k =  4
k =  3
k =  2
k =  1
k =  0

Basicly, x[i::-1] should be in the for not in the print.

answered Feb 4, 2015 at 23:49

Marcin

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4 Comments

Ricardo Cárdenes Over a year ago

And x[i] > i, as you're there

Marcin Over a year ago

@RicardoCárdenes What do you mean?

SugaKookie Over a year ago

Sorry, I don't understand your comment.

Ricardo Cárdenes Over a year ago

x is a range of integers. In this case there's no sense in using x[i], because i... is x[i]! Same as you're printing k straight, instead of x[k]

GLHF · Accepted Answer · 2015-02-05 00:04:44Z

2

What about just manipulate it with print function?

i = 0
k = 0

while True:
    print (i)
    print (k)
    if 1<k: #tricky part
        print ("\n".join([str(h) for h in range(0,k+1)][::-1]))
    print ("")
    i += 1
    k += 1
    if i == 10:
        break

answered Feb 5, 2015 at 0:04

GLHF

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Comments

Sigve Karolius · Accepted Answer · 2015-02-05 00:00:36Z

0

You are very close. If you are new to the world of python you can take some inspiration from this example where I use list comprehension.

list = lambda k: [ [ i for i in reversed(xrange(j+1)) ] for j in xrange(k+1) ]

Note: If you are using python 3 "xrange" is changed to "range"

Now call:

list(3)

And you'll see that the result is similar to what you are looking for.

[[0], [1, 0], [2, 1, 0], [3, 2, 1, 0]]

answered Feb 5, 2015 at 0:00

Sigve Karolius

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1 Comment

Anthony Pham Over a year ago

The asker wants to print the results, not put in a list. But nice answer!

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