如何比较Python中两个字典列表？

Question

我想知道怎么比较两个字典列表。结果应该是从字典列表B中找出那些不同的项。

举个例子：

ldA = [{'user':"nameA", 'a':7.6, 'b':100.0, 'c':45.5, 'd':48.9},
       {'user':"nameB", 'a':46.7, 'b':67.3, 'c':0.0, 'd':5.5}]


ldB =[{'user':"nameA", 'a':7.6, 'b':99.9, 'c':45.5, 'd':43.7},
      {'user':"nameB", 'a':67.7, 'b':67.3, 'c':1.1, 'd':5.5},
      {'user':"nameC", 'a':89.9, 'b':77.3, 'c':2.2, 'd':6.5}]

在这里，我想比较ldA和ldB。它应该输出下面的结果。

ldB -> {user:"nameA",  b:99.9, d:43.7}
ldB -> {user:"nameB",  a:67.7, c:1.1 }
ldb -> {user:"nameC", a:89.9, b:77.3, c:2.2, d:6.5}

我看过下面的链接，但那里只返回了名字，而我想要的是名字和对应的值，就像上面那样。

比较字典列表以匹配列表并检测Python中的值变化

Answer 1

我假设这两个列表里的dict是按相同的顺序排列的。

在这个假设下，你可以使用下面的代码：

def diffs(L1, L2):
    answer = []
    for i, d1 in enumerate(L1):
        d = {}
        d2 = L2[i]
        for key in d1:
            if key not in d1:
                print key, "is in d1 but not in d2"
            elif d1[key] != d2[key]:
                d[key] = d2[key]
        answer.append(d)
    return answer

这段代码还没有测试过。如果有错误，请留言告诉我，我会修正的。

Answer 2

我的方法是：先建立一个查找表，里面包含要排除的值，然后再从每个列表中找出合适的值进行排除，得到最终的结果。

lookup = dict((x['user'], dict(x)) for x in ldA)
# 'dict(x)' is used here to make a copy
for v in lookup.values(): del v['user']

result = [
    dict(
        (k, v)
        for (k, v) in item.items()
        if item['user'] not in lookup or lookup[item['user']].get(k, v) == v
    )
    for item in ldB
]

不过，你需要知道，像这样比较浮点数的值是不能完全依赖的。

Answer 3

对于一个通用的解决方案，可以考虑以下内容。即使用户在列表中的顺序不一致，它也能正确地进行差异比较。

def dict_diff ( merge, lhs, rhs ):
    """Generic dictionary difference."""
    diff = {}
    for key in lhs.keys():
          # auto-merge for missing key on right-hand-side.
        if (key not in rhs):
            diff[key] = lhs[key]
          # on collision, invoke custom merge function.
        elif (lhs[key] != rhs[key]):
            diff[key] = merge(lhs[key], rhs[key])
    for key in rhs.keys():
          # auto-merge for missing key on left-hand-side.
        if (key not not lhs):
            diff[key] = rhs[key]
    return diff

def user_diff ( lhs, rhs ):
    """Merge dictionaries using value from right-hand-side on conflict."""
    merge = lambda l,r: r
    return dict_diff(merge, lhs, rhs)

import copy

def push ( x, k, v ):
    """Returns copy of dict `x` with key `k` set to `v`."""
    x = copy.copy(x); x[k] = v; return x

def pop ( x, k ):
    """Returns copy of dict `x` without key `k`."""
    x = copy.copy(x); del x[k]; return x

def special_diff ( lhs, rhs, k ):
      # transform list of dicts into 2 levels of dicts, 1st level index by k.
    lhs = dict([(D[k],pop(D,k)) for D in lhs])
    rhs = dict([(D[k],pop(D,k)) for D in rhs])
      # diff at the 1st level.
    c = dict_diff(user_diff, lhs, rhs)
      # transform to back to initial format.
    return [push(D,k,K) for (K,D) in c.items()]

接下来，你可以检查这个解决方案：

ldA = [{'user':"nameA", 'a':7.6, 'b':100.0, 'c':45.5, 'd':48.9},
       {'user':"nameB", 'a':46.7, 'b':67.3, 'c':0.0, 'd':5.5}]
ldB =[{'user':"nameA", 'a':7.6, 'b':99.9, 'c':45.5, 'd':43.7},
      {'user':"nameB", 'a':67.7, 'b':67.3, 'c':1.1, 'd':5.5},
      {'user':"nameC", 'a':89.9, 'b':77.3, 'c':2.2, 'd':6.5}]
import pprint
if __name__ == '__main__':
    pprint.pprint(special_diff(ldA, ldB, 'user'))