Create list of strings with fixed prefix based on size of a list

Question

I want to create a list of strings with a fixed prefix where the suffix is based on the size of a list, by using python 3.3.2.

For example I have a list elements with len(elements) equal to 3. The output should be a list output = ['prefix_1','prefix_2','prefix_3']

I can do this by using a loop:

elements = ['elem1','elem2','elem3']
output = []
for i in range(len(elements)):
    output.append('prefix_'+str((i+1)))

This works but seems a bit... unpythonic to me. Is there a more pythonic way to do this? For example with a list comprehension?

falsetru · Accepted Answer · 2013-11-12 08:43:22Z

5

Using enumerate, str.format:

>>> elements = ['elem1','elem2','elem3']
>>> output = ['prefix_{}'.format(i) for i, elem in enumerate(elements, 1)]
>>> output
['prefix_1', 'prefix_2', 'prefix_3']

answered Nov 12, 2013 at 8:43

falsetru

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Manoj Awasthi · Accepted Answer · 2013-11-12 08:43:42Z

1

o = ["prefix_" + str(i+1) for i in range(len(elements))]

answered Nov 12, 2013 at 8:43

Manoj Awasthi

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Comments

Feanor · Accepted Answer · 2013-11-12 08:44:21Z

1

output=['prefix_'+str(i+1) for i in range(len(['elem1','elem2','elem3']))]

answered Nov 12, 2013 at 8:44

Feanor

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