Question

如何比较两个dict列表？结果应该是dict B列表中的奇数。

示例：

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}]

这里我想比较ldA和ldB。它应该打印下面的输出。

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}

我已经浏览了下面的链接，但是它只返回名称，但我想要名称和价值如上所述。

List of Dicts comparision to match between lists and detect value changes in Python

Answer 1

对于一般解决方案，请考虑以下事项。即使用户在列表中出现故障，它也会正确区分。

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 (not rhs.has_key(key)):
            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 (not lhs.has_key(key)):
            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()]

然后，您可以检查解决方案：

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'))

Answer 2

我的方法：根据要排除的值的ldA构建查找，然后确定从ldB中排除每个列表中的适当值的结果。

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
]

You should, however, be aware that comparing floating-point values like that can't be relied upon

Answer 3

我将假设相应的dict在两个列表中的顺序相同。

根据该假设，您可以使用以下代码：

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

未测试。请评论是否有错误，我会解决它们

Answer 4

另一个解决方案有点奇怪（对不起，如果我错过了一些东西）但它也允许你配置你自己的相等检查（你只需要为此修改isEqual lambda）以及给你两个不同的选项如何处理在密钥不同的情况下：

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}]

ldA.extend((ldB.pop() for i in xrange(len(ldB)))) # get the only one list here

output = []

isEqual = lambda x,y: x != y # add your custom equality check here, for example rounding values before comparison and so on

while len(ldA) > 0: # iterate through list
    row = ldA.pop(0) # get the first element in list and remove it from list
    for i, srow in enumerate(ldA):
        if row['user'] != srow['user']:
            continue
        res = {'user': srow['user']} #
        # next line will ignore all keys of srow which are not in row 
        res.update(dict((key,val) for key,val in ldA.pop(i).iteritems() if key in row and isEqual(val, row[key])))
        # next line will include the srow.key and srow.value into the results even in a case when there is no such pair in a row
        #res.update(dict(filter(lambda d: isEqual(d[1], row[d[0]]) if d[0] in row else True ,ldA.pop(i).items())))
        output.append(res)
        break
    else:
        output.append(row)

print output

Answer 5

这肯定会从您的示例数据中做出一些假设，主要是ldA中的用户不会在ldB中，如果这是一个无效的假设，请告诉我。

您可以将其称为dict_diff(ldA, ldB, user)。

def dict_diff(ldA, ldB, key):
    for i, dA in enumerate(ldA):
        d = {key: dA[key]}
        d.update(dict((k, v) for k, v in ldB[i].items() if v != dA[k]))
        print "ldB -> " + str(d)
    for dB in ldB[i+1:]:
        print "ldB -> " + str(dB)

Answer 6

我前一段时间写了this tool，它目前可以处理嵌套列表，词组和集合。为您提供更简洁的输出（.中的. > i:1 > 'c'引用顶级，i:1引用要比较的列表的索引1）：

compare(ldA, ldB)
. > i:0 > 'b' dict value is different:
100.0
99.9

. > i:0 > 'd' dict value is different:
48.9
43.7

. > i:1 > 'a' dict value is different:
46.7
67.7

. > i:1 > 'c' dict value is different:
0.0
1.1

. lists differed at positions: 2
['<not present>']
[{'c': 2.2, 'd': 6.5, 'a': 89.9, 'user': 'nameC', 'b': 77.3}]

如何在Python中比较两个dicts列表？

6 个答案: