4

This is a problem I've come across a lot lately. Google doesn't seem to have an answer so I bring it to the good people of stack overflow.

I am looking for a simple way to populate a list with the output of a function. Something like this:

fill(random.random(), 3) #=> [0.04095623, 0.39761869, 0.46227642]

Here are other ways I've found to do this. But I'm not really happy with them, as they seem inefficient.

results = []
for x in xrange(3): results.append(random.random())
#results => [0.04095623, 0.39761869, 0.46227642]

and

map(lambda x: random.random(), [None] * 3)
#=> [0.04095623, 0.39761869, 0.46227642]

Suggestions?

Thanks for all the answers. I knew there was a more python-esque way.

And to the efficiency questions...

$ python --version
Python 2.7.1+
$ python -m timeit "import random" "map(lambda x: random.random(), [None] * 3)"
1000000 loops, best of 3: 1.65 usec per loop
$ python -m timeit "import random" "results = []" "for x in xrange(3): results.append(random.random())"
1000000 loops, best of 3: 1.41 usec per loop
$ python -m timeit "import random" "[random.random() for x in xrange(3)]"
1000000 loops, best of 3: 1.09 usec per loop
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5 回答 5

12

How about a list comprehension?

[random.random() for x in xrange(3)]

Also, in many cases, you need the values just once. In these cases, a generator expression which computes the values just-in-time and does not require a memory allocation is preferable:

results = (random.random() for x in xrange(3))
for r in results:
   ...
# results is "used up" now.
# We could have used results_list = list(results) to convert the generator

By the way, in Python 3.x, xrange has been replaced by range. In Python 2.x, range allocates the memory and calculates all values beforehand (like a list comprehension), whereas xrange calculates the values just-in-time and does not allocate memory (it's a generator).

于 2012-01-02T18:21:22.303 回答
2

why do you think they are inefficient?

There is another way to do it,a list-comprehension 

listt= [random.random() for i in range(3)]
于 2012-01-02T18:22:13.593 回答
0 
   list =  [random.random() for i in xrange(3)]
   list =  [random.random() for i in [0]*3]
   list =  [i() for i in [random.random]*3]

Or :

  fill =lambda f,n: [f() for i in xrange(n)]
  fill(random.random , 3 ) #=> [0.04095623, 0.39761869, 0.46227642]
于 2012-01-02T18:20:30.817 回答
0

更通用的东西...

from random import random

fill = lambda func, num: [func() for x in xrange(num)]
# for generating tuples:
fill = lambda func, num: (func() for x in xrange(num))


# then just call:
fill(random, 4)
# or...
fill(lambda : 1+2*random(), 4)
于 2012-01-02T18:28:10.277 回答
0

列表理解可能是最清楚的，但对于itertools爱好者来说：

>>> list(itertools.islice(iter(random.random, None), 3))
[0.42565379345946064, 0.41754360645917354, 0.797286438646947]

快速检查timeit表明 itertools 版本对于 10 多个项目的速度稍快一些，但仍然可以使用您认为最清楚的任何内容：

C:\Python32>python lib\timeit.py -s "import random, itertools" "list(itertools.islice(iter(random.random, None), 10))"
100000 loops, best of 3: 2.93 usec per loop

C:\Python32>python lib\timeit.py -s "import random, itertools" "[random.random() for _ in range(10)]"
100000 loops, best of 3: 3.19 usec per loop
于 2012-01-02T19:03:23.033 回答