For Loop generate sequence

Question

Need help,

I expected the output like this

1.1.1.1 is server0
1.1.1.2 is server1
1.1.1.3 is server2
1.1.1.4 is server3

But I don't get the output like I expected. Here's my code

IP = ['1.1.1.1', '1.1.1.2', '1.1.1.3', '1.1.1.4']
for ip in IP:
    for i in range(len(IP)):
        print(ip + ' is ' + 'server'+ str(i))

Output:

1.1.1.1 is server0
1.1.1.1 is server1
1.1.1.1 is server2
1.1.1.1 is server3
1.1.1.2 is server0
1.1.1.2 is server1
1.1.1.2 is server2
1.1.1.2 is server3
1.1.1.3 is server0
1.1.1.3 is server1
1.1.1.3 is server2
1.1.1.3 is server3
1.1.1.4 is server0
1.1.1.4 is server1
1.1.1.4 is server2
1.1.1.4 is server3

Can someone help me?

Use enumerate : `for i, ip in enumerate(IP): print(ip + ' is ' + 'server'+ str(i))` — eyllanesc, Jun 22 '18 at 04:46
I have rolled back your last edit. Don't fundamentally alter your question after it already received an answer. If you have a new question, then post a new question. — Mark Rotteveel, Jun 22 '18 at 12:04

score 0 · Accepted Answer · answered Jun 22 '18 at 04:46

0

You don't need two loops here. Simply write:

for i in range(len(IP)):
            print(IP[i] + ' is ' + 'server'+ str(i))

Output:

1.1.1.1 is server0
1.1.1.2 is server1
1.1.1.3 is server2
1.1.1.4 is server3

answered Jun 22 '18 at 04:46

Taohidul Islam

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`for i, ip in enumerate(IP): print(ip + ' is server' + str(i))` Is a nice way to do it. – RoadRunner Jun 22 '18 at 05:54
Yeah, but i think time complexity of both of them are almost same. – Taohidul Islam Jun 22 '18 at 05:56
Time complexity shouldn't be an issue for a simple loop. When you want to use the index and the item during iteration, `enumerate()` is the most pythonic choice. – RoadRunner Jun 22 '18 at 06:01
Yeah, thanks for your point – Taohidul Islam Jun 22 '18 at 06:18

For Loop generate sequence

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