如何返回两个二维列表中的匹配项和不匹配项

Question

我有两个 2d 列表，超过 200 行。我想在每个子列表中找到 df1 和 df2 中的匹配项，并且仅当 df1、df2 中的值匹配时才返回不匹配和匹配值。

这是一个简短的例子

 df1 = [[1, 7, 3, 5], [5, 5, 14, 10]]

 df2 = [[1, 17, 3, 5], [34, 14, 74], [34, 3, 87], [25, 14, 10]]

想要的结果

 no_match = [[17],[34,87],[34,74],[25]]

 match = [[1,3,5],[3],[14],[14,10]]   
  

Answer 1

试试这个：

df1 = [[1, 7, 3, 5], [5, 5, 14, 10]]
df2 = [[1, 17, 3, 5], [34, 14, 74], [34, 3, 87], [25, 14, 10]]

no_match = []
match = []

for sublist in df2:
    sublist_match = []
    sublist_no_match = []
    for item in sublist:
        for sublist_df1 in df1:
            if item in sublist_df1:
                sublist_match.append(item)
                break
        else:
            sublist_no_match.append(item)
    no_match.append(sublist_no_match)
    match.append(sublist_match)

print("no_match =", no_match)
print("match =", match)

输出：

no_match = [[17], [34, 74], [34, 87], [25]]
match = [[1, 3, 5], [14], [3], [14, 10]]

Answer 2

您可以简单地展平并从

df1

中删除重复项，然后迭代

df2

确定当前项目是否位于展平的

df1

中。

# data
df1 = [[1, 7, 3, 5], [5, 5, 14, 10]]
df2 = [[1, 17, 3, 5], [34, 14, 74], [34, 3, 87], [25, 14, 10]]

# flatten df1 and remove duplicates
s1 = set([i for l in df1 for i in l])

# init containers
match    = []
no_match = []

for data in df2:
    # prime containers for this iteration of data
    match.append([])
    no_match.append([])
    
    # append item to the proper container
    for item in data:
        (no_match, match)[item in s1][-1].append(item)
        
# print results
print('match:', match)
print('no match:', no_match)