如何比较多个列表元素的位置？

Question

我有多个字符串列表。 我需要将列表中每个位置的内容与其他列表中相同位置的内容进行比较。 然后计算多少个相同（不包括“ 0”）。 列表的长度相同。 这是一个例子：

list1 = ["a", "b", "c", "f", "0"]
list2 = ["a", "b", "e", "f", "0"]
list3 = ["a", "0", "c", "f", "0"]

因此，在所有列表中只有相同位置的"a"和"f" （不计“ 0”），那么我需要一个等于2的输出。

我尝试了以下策略，但由于它仅将两个值进行两个比较，因此失败了，并且可以修改np.sum以便比较所有列表。

All_lists包含所有列表。

def scores(All_lists):
    score_lin = []
    score_ = 0
    for j in range(len(All_lists[0])):
        score_lin.append(
            np.sum(
                All_lists[j] == All_lists[j+1])
                - min(
                    np.count_nonzero(All_lists[j] == '0'),                 
                    np.count_nonzero(All_rg_lists[j+1] == '0')
                )
            )

    score_lin = [
        item 
        for item in score_lin 
        if item > 0
    ]
    score_ = sum(score_lin)

    return score_

编辑：用于列表的列表。

list_ = [
    [
        ["a", "b", "c", "f", 0], 
        ["a", "b", "e", "f", 0], 
        ["a", 0, "c", "f", 0]
    ], 
    [
        ["b", "x", "c", "f", 0],  
        ["a", "b", "c", "f", 0],  
        ["a", "b", "c", "f", 0]
    ],
]

考虑列表“ list”的列表。 我需要比较list[0][0] = ["a","b","c","f",0]与list[1][0] = ["b","x","c","f",0]并计算两个列表中相似元素的数量，然后对list[0][1]和list[1][1]...在此示例中，预期输出是8。

nb：列表的长度可以大于3。

Answer 1

您可以使用zip()将这三个列表压缩在一起，并可以使用set()查找重复项：

list1 = ["a", "b", "c", "f", "0"]
list2 = ["a", "b", "e", "f", "0"]
list3 = ["a", "0", "c", "f", "0"]

print(
    sum(
        len(set(i))==1 and i[0]!='0' 
        for i in zip(list1, list2, list3)
    )
)

印刷品：

2

编辑：对于列表中的列表：

l = [
    [
        ['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's1', 0, 0, 0, 0, 0]
    ],
    [
        ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
        ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
        ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
        ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
        ['s2', 's1', 's3', 's3', 0, 0, 0, 0, 0, 0]
    ],
    [
        ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0],
        ['s2', 's1', 's2', 's2', 's3', 's3', 0, 0, 0, 0]
    ]
]

from pprint import pprint

total_sum = 0
for item in zip(*l):
    pprint(item)
    s = sum(
        len(set(i))==1 and i[0]!=0
        for i in zip(*item)
    )
    total_sum += s
    print(s)
    print('*' * 80)

print('Total sum =', total_sum)

印刷品：

(['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
 ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
 ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0])
2
********************************************************************************
(['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
 ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
 ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0])
2
********************************************************************************
(['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
 ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
 ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0])
2
********************************************************************************
(['s2', 's1', 's2', 's2', 's2', 's1', 0, 0, 0, 0],
 ['s2', 's1', 's1', 's3', 's3', 0, 0, 0, 0, 0],
 ['s2', 's1', 's2', 's2', 's2', 0, 0, 0, 0, 0])
2
********************************************************************************
(['s2', 's1', 's2', 's2', 's1', 0, 0, 0, 0, 0],
 ['s2', 's1', 's3', 's3', 0, 0, 0, 0, 0, 0],
 ['s2', 's1', 's2', 's2', 's3', 's3', 0, 0, 0, 0])
2
********************************************************************************
Total sum = 10

Answer 2

如果您想要一个可以接受任意数量列表的通用函数，则可以使用它，

def score(All_lists):
    z = zip(*All_lists) # zip lists together to make list of tuples
    z = [t for t in z if '0' not in t] # remove all '0' entries
    score_list = [1 for t in z if len(set(t))==1]
    return sum(score_list)

对于这些输入列表，

list1=["a","b","c","f","0"]
list2=["a","b","e","f","0"]
list3=["a","0","c","f","0"]
All_lists = [list1, list2, list3]

调用score(All_lists)返回2 。

Answer 3

您可以这样进行：

list1=["a","b","c","f","0"]
list2=["a","b","e","f","0"]
list3=["a","0","c","f","0"]

print(len([len(set(x)) for x in list(zip(list1,list2,list3)) if len(set(x)) == 1 and x[0] != '0']))

输出：

 2

Answer 4

关于什么 ：

sum(a == b == c  for a, b, c  in zip(list1, list2, list3) if a != '0')