Question

我有X个列表，例如：

[potato, pie]
[chicken,chicken,pie,donkey,potato,potato]

我想检查一个术语出现在多少个列表中：

例如：

使用以上两个列表，我希望输出为：

(potato,2)
(pie,2)
(chicken,1) - chicken is only one because it appears only in list two, not in list one.
(donkey,1)

我的尝试，但是我做错了，并且变得困惑，甚至我采取了正确的方法：

x = ['potato', 'pie']
z = ['chicken','chicken','pie','donkey','potato','potato']
list_final = x + z

dict_final = {}

for item in list_final:
    if item in dict_final.keys():
        dict_final.update({item:(dict_final.get(item) + 1)})
    else:
        dict_final.update({item:1})


print(dict_final)

我尝试过这个，但这只是计算它出现在列表中的所有时间：

{'potato': 3, 'pie': 2, 'chicken': 2, 'donkey': 1}

但是我想得到：

{'potato': 2, 'pie': 2, 'chicken': 1, 'donkey': 1}

Answer 1

如果您使其更具功能性）


from collections import Counter
from functools import reduce

x = ['potato', 'pie']
y = ['chicken','chicken','pie','donkey','potato','potato']
all_lists = [x, y]
dict(Counter(reduce(lambda x, y: x + list(set(y)), all_lists, [])).most_common())
# {'potato': 2, 'pie': 2, 'donkey': 1, 'chicken': 1}

Answer 2

您应该遍历项目列表的set，以免计数两次。除此之外，您还必须遍历列表中的所有和项：

x = ['potato', 'pie']
z = ['chicken','chicken','pie','donkey','potato','potato']
items = set(x + z)
dict_final = {}

for i in items:
    for lst in [x, z]:
        if i in lst:
            dict_final[i] = dict_final.get(i, 0) + 1

# {'chicken': 1, 'donkey': 1, 'potato': 2, 'pie': 2}

要创建元组列表，您可以

list_final = []
for i in items:
    cntr = sum([i in lst for lst in (x, z)])
    list_final.append((i, cntr))

...以及对oneliners的粉丝：

list_final = [(i, sum([i in lst for lst in (x, z)])) for i in items]
dict_final = {i: sum([i in lst for lst in (x, z)]) for i in items}

Answer 3

仅出于兴趣的考虑：这里是目前为止可用的解决方案的时间：

WITH cte AS (
    SELECT t.*
         , CASE WHEN "IN"  IS NOT NULL THEN COUNT("IN")  OVER (ORDER BY "TIMESTAMP") END AS rn1
         , CASE WHEN "OUT" IS NOT NULL THEN COUNT("OUT") OVER (ORDER BY "TIMESTAMP") END AS rn2
    FROM t
)
SELECT cte.*
     , LEAD("OUT") OVER (ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST) AS NEXT_OUT
FROM cte
ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST

计时代码：

1. double for loop dict    :   8.1e-06
2. for loop LC into dict   :   1.32e-05
3. LC into LC              :   1.37e-05
4. for loop LC into list   :   1.41e-05
5. LC into DC              :   1.41e-05
6. Counter/reduce          :   2.47e-05
7. Counter/reduce into dict:   2.64e-05

一个术语出现在多少个列表中

3 个答案: