在一定范围内查找两个列表中的元素

So I have two lists L1 is fomatted like this:

所以我有两个列表L1是这样的：

L1 = ['12:55:35.87', '12:55:35.70', ...]
L2 = ['12:55:35.53', '12:55:35.30', ...]

I am trying to find pairs in both list that start with the same 4 characters i.e. xx:x and then return the indexes of the pairs for each list

我试图在两个列表中找到以相同的4个字符开头的对，即xx：x然后返回每个列表的对的索引

So far I have:

到目前为止我有：

for pair1 in L1:
    for pair2 in L2:
        if pair1[:4] in pair2:
            print(L1.index(pair1))

This doesn't seem to return the correct indexes and it obviously doesn't return the index of the second list. Any help would be greatly appreciated.

这似乎没有返回正确的索引，它显然不返回第二个列表的索引。任何帮助将不胜感激。

6 个解决方案

#1

Here's how to make your code work. Keep in mind this is a naive solution, there are faster way to solve this if your lists are big. Runtime here is O(n^2) but this could be solved in linear time.

以下是如何使您的代码工作。请记住，这是一个天真的解决方案，如果您的列表很大，有更快的方法来解决这个问题。这里的运行时间是O（n ^ 2），但这可以在线性时间内解决。

for i,pair1 in enumerate(L1):
    for j,pair2 in enumerate(L2):
        if pair1[:4] == pair2[:4]:
            print("list1: %s , list2: %s" % (i,j))

Update: for future visitors here's an average linear time solution:

更新：对于未来的访客，这是一个平均线性时间解决方案：

from collections import defaultdict
l1_map = defaultdict([])

for i,val in enumerate(L1):
    prefix = val[:4]
    l1_map[prefix].append(i)


for j,val in enumerate(L2):
     prefix = val[:4]
     for l1 in l1_map[prefix]:
        print("list1: %s , list2: %s" % (l1,j))

#2

Because OP lists seem to have lots of repeated "firsts 4 characters", I would do something like the following:

因为OP列表似乎有很多重复的“第一个4个字符”，我会做类似以下的事情：

indices = {}
for i, entry in enumerate(L1):
    indices.setdefault(entry[:4], [])
    indices[entry[:4]].append("L1-{}".format(i))
    if L2[i][:4] in indices:
        indices[L2[i][:4]].append("L2-{}".format(i))

Then you can access your repeated entries as:

然后您可以访问重复的条目：

for key in indices:
    print(key, indices[key])

This is better than O(n^2).

这比O（n ^ 2）好。

edit: as someone pointed out in the comments this is assuming that the lists do share the same length.

编辑：正如有人在评论中指出的那样，假设列表的长度相同。

In case they don't, assume L2 is larger than L1, then after doing the above you can do:

如果他们不这样做，假设L2大于L1，那么在完成上述操作后，您可以：

for j, entry in enumerate(L2[i+1:]):
    indices.setdefault(entry[:4], [])
    indices[entry[:4]].append("L2-{}".format(j))

If L2 is shorter than L1 just change the variables names in the code shown.

如果L2比L1短，只需更改所示代码中的变量名称即可。

#3

You can use itertools.product to loop the Cartesian product.

您可以使用itertools.product循环Cartesian产品。

from itertools import product

L1 = ['12:55:35.87', '12:55:35.70']
L2 = ['12:55:35.53', '12:45:35.30']

res = [(i, j) for (i, x), (j, y) in 
       product(enumerate(L1), enumerate(L2)) 
       if x[:4] == y[:4]]

# [(0, 0), (1, 0)]

#4

Use the range() or enumerate() function in the for-loops to provide you loop index.

在for循环中使用range（）或enumerate（）函数为您提供循环索引。

For example, using the range() function:

例如，使用range（）函数：

for x in range(len(L1)):
   for y in range(len(L2)):
       if L1[x][:4] == L2[y][:4]:
           print(x, y)

#5

enumerate is great for things like this.

枚举对于这样的事情很有用。

indexes = []
for index1, pair1 in enumerate(L1):
    pair1_slice = pair1[:4] 
    for index2, pair2 in enumerate(L2):        
        if pair1_slice == pair2[:4]:
            indexes.append([index1, index2])
            print(index1, index2)

#6

I think the enumerate function is what you're looking for!

我认为枚举功能正是你要找的！

L1 = ['12:55:35.87', '12:55:35.70', 'spam']
L2 = ['12:55:35.53', 'eggs', '12:55:35.30']

idxs = []

for idx1, pair1 in enumerate(L1):
    for idx2, pair2 in enumerate(L2):
        if pair1[:4] == pair2[:4]:
            idxs.append((idx1, idx2))

print(idxs)

Output

产量

[(0, 0), (0, 2), (1, 0), (1, 2)]

#1