Question

我的字典看起来像这样：

entryDir = {'cities': [['US', 'AU']], 'countries': [['DK', 'US']]}

如果每个列表中不同键的值匹配，我试图弄清楚如何打印确认。在伪代码中：

for each key:
    # {cities} {countries}
    if a value for that key matches the value of another key:
        #cities:US == countries:US
        print key1:value1,key2:value2,"match found" 
        #cities:US,countries:US,"match found
    if they don't match:
        #cities:AU == countries:DK
        print key1:value2,key2:value1,"no match"
        cities:AU,countries:DK,"no match found"

到目前为止我所拥有的是：

for key1 in entryDir:
    for key2 in entryDir:
        if key1 != key2:                        
            if entryDir[key1][0] == entryDir[key2][0]:
                print entryDir[key1][0],entryDir[key2][0],"match found"
            if entryDir[key1][0] != entryDir[key2][0]:
                print entryDir[key2][0],entryDir[key2][0],"no match found"

但这并不是正确匹配或不匹配。

Answer 1

您还将键值与其自身值进行比较。请尝试改为：

for key1 in entryDir:
    for key2 in entryDir:
        if key1 != key2:
            if entryDir[key1][0] == entryDir[key2][0]:
                print entryDir[key1][0],entryDir[key2][0],"match found"
            else:
                print entryDir[key2][0],entryDir[key2][0],"no match found"

修改

如果您希望每个键有多个值，则需要这样做：

for key1 in entryDir: for key2 in entryDir: if key1 != key2: for val in entryDir[key1]: if val in entryDir[key2] print val, "match found"

请注意根据输入的大小计算需要计算的（可能）可怕时间。

Answer 2

我已经解决了。我可以使用entryDir.keys（）生成的数组的索引作为比较entryDir中的值的索引。

所以，而不是：

for key1 in entryDir:
    for key2 in entryDir:
        if key1 != key2:
            for val in entryDir[key1]:
                if val in entryDir[key2]
                    print val, "match found"

我们可以做到：

keyList=entryDir.keys()
if keyList[0] != keyList[1]:                                            
    if entryDir[keyList[0]][0] == entryDir[keyList[1]][0]:
        print entryDir[keyList[0]][0], "match found"
    else:
        print entryDir[keyList[0]][0],entryDir[keyList[1]][0], "no match"

迭代列表的dict并返回具有匹配值

2 个答案: