Is there a way to append elements to a list randomly, built in function
ex:
def random_append():
lst = ['a']
lst.append('b')
lst.append('c')
lst.append(开发者_C百科'd')
lst.append('e')
return print lst
this will out put ['a', 'b', 'c', 'd', 'e']
But I want it to add elements randomly and out put something like this:
['b', 'd', 'b', 'e', 'c']
And yes there's a function random.shuffle() but it shuffles a list at once which I don't require, I just want to perform random inserts only.
If you need to perform single insert in a random position then the already given trivial example works:
from random import randrange
def random_insert(lst, item):
lst.insert(randrange(len(lst)+1), item)
However if you need to insert k items to a list of length n then using the previously given function is O(n*k + k**2) complexity. However inserting multiple items can be done in linear time O(n+k) if you calculate the target positions ahead of time and rewrite the input list in one go:
from random import sample
def random_insert_seq(lst, seq):
insert_locations = sample(xrange(len(lst) + len(seq)), len(seq))
inserts = dict(zip(insert_locations, seq))
input = iter(lst)
lst[:] = [inserts[pos] if pos in inserts else next(input)
for pos in xrange(len(lst) + len(seq))]
If there is supposed to be exactly one of each item
>>> from random import randint
>>> a=[]
>>> for x in "abcde":
... a.insert(randint(0,len(a)),x)
...
>>> a
['b', 'a', 'd', 'c', 'e']
If you are allowing duplicates (as the output indicates)
>>> from random import choice
>>> a=[choice("abcde") for x in range(5)]
>>> a
['a', 'b', 'd', 'b', 'a']
random.shuffle
is probably the best tool for the job. It is simple, obvious, and well-named—it's probably more readable than the other suggestions you will get. Additionally, using it is O(n), but using insert
(an O(n) operation) n times is quadratic.
from random import choice
n=10
seq=['a','b','c','d']
rstr=[choice(seq) for i in range(n)]
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