Python nested for loop counting downwards

Question

I have an assignment to create a counting down for loop. The instructions are to get the results to be as follows:

I'm probably making this to be more difficult than it actually is, but I'm quite new to Python.

So far, my script is as follows:

def print_figure():
    for i in range(5, 0, -1):
        for count in range(i):
            print(12345, end='')
        print()

Which is resulting in:

1234512345123451234512345
12345123451234512345
123451234512345
1234512345
12345

I'm not asking for an answer, just a pointer in the right direction of where I can make a fix.

Martijn Pieters · Accepted Answer · 2013-11-08 18:59:45Z

1

You were almost there; you are printing the number 12345 over and over again, instead of using count, which ranges from 0 through to 4 in the first loop, 0 to 3 in the second, etc.

You want to print count itself, but either add one or adjust the range in which it loops to run from 1 to i + 1:

def print_figure():
    for i in range(5, 0, -1):
        for count in range(1, i + 1):
            print(count, end='')
        print()

or use:

def print_figure():
    for i in range(5, 0, -1):
        for count in range(i):
            print(count + 1, end='')
        print()

Demo:

>>> def print_figure():
...     for i in range(5, 0, -1):
...         for count in range(1, i + 1):
...             print(count, end='')
...         print()
... 
>>> print_figure()
12345
1234
123
12
1

The other approach would be to use a string '12345' and indexing into that; '12345'[0] is '1', etc:

def print_figure():
    for i in range(5, 0, -1):
        for count in range(i):
            print('12345'[count], end='')
        print()

but then you can just use slicing and get rid of the nested loop altogether:

def print_figure():
    for i in range(5, 0, -1):
        print('12345'[:i])

because '12345'[:3] returns the first 3 characters of the string, printing 123.

edited Nov 8, 2013 at 18:59

answered Nov 8, 2013 at 18:49

Martijn Pieters

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2 Comments

Izkata Over a year ago

I have the feeling the OP was halfway between that and "12345"[count]

heb-NR Over a year ago

This was great. Thanks guys. I was trying to explain it to someone earlier while I was out and the idea of an index and a slice came to mind. I just wasn't sure how to input that. Fantastic! Thanks again!

Raiyan · Accepted Answer · 2013-11-08 19:16:16Z

1

replace the last 2 lines :

        print(12345, end='')
    print()

with

        print(count),
    print ''

answered Nov 8, 2013 at 19:16

Raiyan

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Comments

njzk2 · Accepted Answer · 2013-11-08 19:21:39Z

0

A more generic approach would be to take a string a remove the last item at every iteration :

value = '12345'
while value:
    print value
    value = value[:-1]

The upside is that you can easily replace 12345 with any string, of any length.

answered Nov 8, 2013 at 19:21

njzk2

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