Index of List in python

Question

I have this codes

def display(s1,s2):
    l1 = list(s1)
    l2 = list(s2)
    l3 = [None]*10
    for i in range(10):
        for j in range(10):
            if i==l2[j]:
                l3[j] = l1[i]
    return l3

print display('3941068257', '1234567890')

Example: 3 in position 0 of list1; 0 in position 9 of list2 => we will display 3 in position 9 in new list called l3 and so on ....

so program supposes to display like that 9410682573 but it still display None none ......

My compiler doesn't have debugger so I don't know how to find out. Anyone can help?

Tadeck · Accepted Answer · 2012-05-23 02:13:18Z

5

You are comparing string (one-char element of l2) with integer. It always fails.

It is because the below happens:

l2 = ['3','9','4','1','0','6','8','2','5','7']  # when you do "l2 = list(s2)"
for i in [0,1,2,3,4,5,6,7,8,9]:  # when you do "for i in raange(10)"

The types just don't match.

Instead do this:

def display(s1,s2):
    l1 = list(s1)
    l2 = list(s2)
    l3 = [None]*10
    for i in range(10):
        for j in range(10):
            if str(i) == l2[j]:  # <-- change is here
                l3[j] = l1[i]
    return l3

print display('3941068257', '1234567890')

answered May 23, 2012 at 2:13

Tadeck

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1 Comment

Tadeck Over a year ago

@LongBodie: To display a list as one string just join it like that: ''.join(l3).

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