CODEWITHSUNDEEP
pythonlist
Lee
Lee

Reputation: 31040

How can I increase the size of, and pad, a python list?

Say I have this list:

[1,2,3,4]

and I want:

[1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4]

What is the best way of doing this?

My current method is to create a new list:

x = [1,2,3,4]
y = [[n]*4 for n in x]

This gives:

[[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4]]

Which seems close but no cigar... Can anyone help?

Upvotes: 3

Views: 5312

Answers (5)

Sayak Sen
Sayak Sen

Reputation: 11

Here is how I would have done.

x = [1,2,3,4]
y = x*4
y.sort()

Easy and Neat.

Upvotes: 1

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 250931

You can use itertools.chain() or chain.from_iterable() to flatten a list of lists (y in your case) :

In [23]: lis=[1,2,3,4]

In [24]: from itertools import chain

In [31]: y = list(chain(*([n]*4 for n in lis)))

In [32]: y
Out[32]: [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]

Some performance comparisons:

In [25]: x=[1,2,3,4]*1000  #can't use more than 1000 due to limited RAM

In [26]: %timeit list(chain(*([n]*1000 for n in x)))
1 loops, best of 3: 196 ms per loop

In [27]: %timeit list(chain.from_iterable(([n]*1000 for n in x)))
1 loops, best of 3: 177 ms per loop

In [29]: %timeit [n for n in x for _ in xrange(1000)]
1 loops, best of 3: 388 ms per loop

#three more solutions;from @Steven Rumbalski 

In [28]: %timeit [repeated for value in x for repeated in repeat(value,1000)]
1 loops, best of 3: 344 ms per loop

In [30]: %timeit list(chain.from_iterable(izip(*repeat(x,1000))))
1 loops, best of 3: 204 ms per loop

In [31]: %timeit list(chain(*izip(*repeat(x,1000))))
1 loops, best of 3: 238 ms per loop

Upvotes: 2

Pavel Anossov
Pavel Anossov

Reputation: 62908

>>> x = [1,2,3,4]
>>> [n for n in x for _ in range(4)]
    [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4]

 

itertools.repeat is indeed semantically cleaner, thanks, Steven:

from itertools import repeat
[repeated for value in x for repeated in repeat(value, 4)]

Upvotes: 13

nvlass
nvlass

Reputation: 685

You can always use reduce

reduce(lambda x, y: x+y, [[n]*4 for n in x])

Upvotes: 1

Rushy Panchal
Rushy Panchal

Reputation: 17532

Easy method:

y = [n for n in x*4]

This will return [1, 2, 3, 4, 1, 2, 3, 4, ...].

To put it in the order you wanted, you can do x = sorted(x) or y = sorted([n for n in x*4]).

This is similar to y = sorted(x*4).

The best method is to do it how @Pavel Anossov stated:

y = [n for n in x for _ in range(4)]

This is best because it will work for any sequence whereas sorted(x*4) does not keep the order and only works for originally sorted lists.

Upvotes: 0

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