[英]how to create a list sequence in python containing all possible pairs an equal number of times?
Given the letters x,y
is there a way to write a python function that returns a list sequence containing all possible pairs ( [x,x], [x,y], [y,x], [y,y]
) an equal number of times? 给定字母
x,y
,有一种方法可以编写一个python函数,该函数返回一个包含所有可能对( [x,x], [x,y], [y,x], [y,y]
)的列表序列。相等次数?
for example can I write a function that takes the input (eg [x,y]
) and the number of times each possible pair should appear (eg 2
), and then returns, for example, the list sequence: [xxxyyxyyx]
? 例如,我可以编写一个函数来接收输入(例如
[x,y]
)和每个可能的对出现的次数(例如2
),然后返回例如列表序列: [xxxyyxyyx]
吗?
Preferably I would like the function to generate a "random" sequence so it could also return: 优选地,我希望函数生成一个“随机”序列,以便它也可以返回:
[y x y y x x x y y]
In both cases [x,x] [x,y] [y,x]
and [y,y]
appear exactly twice. 在两种情况下,
[x,x] [x,y] [y,x]
和[y,y]
恰好出现两次。
Ideally, I would like to find a solution that also worked with for example 4 letters [x,y,z,w]
. 理想情况下,我想找到一种解决方案,该解决方案也适用于例如4个字母
[x,y,z,w]
。
Based on your edits, the following should conduct the tour to which you were referring. 根据您的编辑,以下内容将指导您所参考的浏览。
import random
import itertools
def generate_sequence(alphabet, num_repeats):
# generate the permutations of the list of size 2
# repeated `num_repeats` times
perms = [(letter1, letter2) for letter1, letter2 in itertools.product(alphabet, alphabet) for i in xrange(num_repeats)]
# save the original length of `perm` for later
perm_len = len(perms)
# randomly choose a starting point and add it to our result
curr_s, curr_e = random.choice(perms)
result = [curr_s, curr_e]
# remove the starting point from the list... on to the next one
perms.remove((curr_s, curr_e))
# while we still have pairs in `perms`...
while len(perms):
# get all possible next pairs if the end of the current pair
# equals the beginning of the next pair
next_vals = [(s,e) for s,e in perms if s == curr_e]
# if there aren't anymore, we may have exhausted the pairs that
# start with `curr_e`, in which case choose a new random pair
if len(next_vals) != 0:
next_s, next_e = random.choice(next_vals)
else:
next_s, next_e = random.choice(perms)
# remove the next pair from `perm` and append it to the `result`
perms.remove((next_s, next_e))
result.append(next_e)
# set the current pair to the next pair and continue iterating...
curr_s, curr_e = next_s, next_e
return result
alphabet = ('x', 'y', 'z')
num_repeats = 2
print generate_sequence(alphabet, num_repeats)
This outputs 这个输出
['z', 'z', 'z', 'x', 'x', 'y', 'z', 'y', 'y', 'z', 'y', 'x', 'y', 'x', 'z', 'x',
'z', 'y', 'x']
With this you can take the most common pairs: 有了这个,您可以选择最常见的对:
>>> from collections import Counter
>>> dict(Counter(zip(a, a[1:])).most_common())
{('z', 'z'): 2, ('z', 'y'): 2, ('x', 'y'): 2, ('z', 'x'): 2, ('y', 'y'): 2, ('x', 'x'): 2, ('y', 'x'): 2, ('x', 'z'): 2, ('y', 'z'): 2}
If you care about the pairs only: 如果您只关心配对:
>>> [t[0] for t in Counter(zip(a, a[1:])).most_common()]
[('z', 'z'), ('z', 'y'), ('x', 'y'), ('z', 'x'), ('x', 'x'), ('y', 'x'), ('x', 'z'), ('y', 'y'), ('y', 'z')]
If you only care which pairs appear 2 times: 如果您只关心哪些对出现2次:
>>> [pair for pair, count in Counter(zip(a, a[1:])).items() if count == 2]
[('z', 'z'), ('z', 'y'), ('x', 'y'), ('z', 'x'), ('x', 'x'), ('y', 'x'), ('x', 'z'), ('y', 'y'), ('y', 'z')]
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