[英]Python group list by same numbers
I need help with python function, which will count for me how many repeated numbers on list, if they are separated by another number.我需要 python function 方面的帮助,如果它们被另一个数字分隔,它将计算列表中有多少重复数字。
nums = [3, 1, 2, 3, 2, 1, 1, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2]
should return应该返回
{
[3]: 2,
[1]: 3,
[2]: 5,
[1, 1]: 2,
[2, 2, 2, 2]: 1
}
You can use itertools.groupby
to collect the consecutive numbers and collections.Counter
to count them:您可以使用
itertools.groupby
收集连续数字和collections.Counter
来统计它们:
nums = [3, 1, 2, 3, 2, 1, 1, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2]
from itertools import groupby
from collections import Counter
dict(Counter(tuple(g) for k,g in groupby(nums)))
NB.注意。 You cannot use lists as dictionary keys, I've used tuples instead
您不能将列表用作字典键,而是使用元组
Output: Output:
{(3,): 2,
(1,): 3,
(2,): 5,
(1, 1): 2,
(2, 2, 2, 2): 1}
Note that you cannot have a list as a key.请注意,您不能将列表作为键。 Instead, we can use a string.
相反,我们可以使用字符串。
This returns all lengths of streaks, not just max and min.这将返回所有长度的条纹,而不仅仅是最大值和最小值。 Say in comments if you would like otherwise.
如果您愿意,请在评论中说。
nums = [3, 1, 2, 3, 2, 1, 1, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2]
results = {}
streak = 0
previous = None
for num in nums:
print(num)
if previous != num:
streak = 0
streak += 1
previous = num
for i in range(1, streak + 1):
results[str(num)*i] = results.get(str(num)*i, 0) + 1
Please comment if you would like an explanation.如果您需要解释,请发表评论。
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