问题 如何在Python中比较两个dicts列表？
我如何比较两个列表 dict？结果应该是dict B列表中的奇数。
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}]
在这里，我想将ldA与ldB进行比较。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}
我已经通过以下链接，但它只返回名称，但我想要如上所述的名称和价值。
列表中的Dicts比较以匹配列表并检测Python中的值更改
7713
2018-06-13 16:48
起源
答案:
对于一般解决方案，请考虑以下内容。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'))
7
2018-06-13 17:33
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
]
但是，您应该知道，不能依赖比较这样的浮点值。
3
2018-06-13 17:25
我将假设相应的 dict两个列表中的顺序相同。
在此假设下，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
未经测试。请评论是否有错误，我会解决它们
2
2018-06-13 17:01
还有一个解决方案有点奇怪（抱歉，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，如何处理以防万一键不同：
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
1
2018-06-13 19:37
这绝对需要您的示例数据中的一些假设，主要是不会有用户 ldA 那些不在 ldB如果这是一个无效的假设，请告诉我。
你会这样称呼它 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)
0
2018-06-13 17:25
我写了 这个工具 不久前，它目前可以处理嵌套列表，词组和集合。给你一个更简洁的输出（ . 在 . > i:1 > 'c' 指的是顶级和 i:1 指的是被比较的列表的索引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}]
0
2018-05-16 22:10