我有一个计算Python中每个键的不同值的问题。蟒蛇字典的唯一值计数
我有一个字典d像
[{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"}, {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"}, {"pqr":"sports"}]
我需要单独打印每每个键的不同值的数目。
这意味着我将要打印
abc 3
xyz 1
pqr 4
请帮助。
谢谢
使用collections.Counter() instance，一些链接在一起：
from collections import Counter
from itertools import chain
counts = Counter(chain.from_iterable(e.keys() for e in d))
这保证了在你的输入列表中有多个键的字典进行正确计数。
演示：
>>> from collections import Counter
>>> from itertools import chain
>>> d = [{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"}, {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"}, {"pqr":"sports"}]
>>> Counter(chain.from_iterable(e.keys() for e in d))Counter({'pqr': 5, 'abc': 3, 'xyz': 1})
或与输入的词典多个键：
>>> d = [{"abc":"movies", 'xyz': 'music', 'pqr': 'music'}, {"abc": "sports", 'pqr': 'movies'}, {"abc": "music", 'pqr': 'sports'}, {"pqr":"news"}, {"pqr":"sports"}]
>>> Counter(chain.from_iterable(e.keys() for e in d)) Counter({'pqr': 5, 'abc': 3, 'xyz': 1})
甲Counter()具有附加的，有益的功能，例如，该目录排序的元件，反向其计数.most_common() method订购：
for key, count in counts.most_common():
print '{}: {}'.format(key, count)
# prints
# 5: pqr
# 3: abc
# 1: xyz
请注意，计数器]（http://docs.python.org/2/library/collections.html#collections.Counter）类是在Python 2.7中引入的。有[backport]（http://code.activestate.com/recipes/576611-counter-class/）。我想你[知道这件事]（Martijn）（http://stackoverflow.com/a/13311111/566644）。 – 2013-05-06 20:14:50
@ LauritzV.Thaulow：以其他方式作为[backport for 2.5 and 2.6]（http://code.activestate.com/recipes/576611-counter-class/）。 – 2013-05-06 20:17:20
...或者你可以在int中使用'defaultdict'。 – 2013-05-06 20:18:23
>>> d = [{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"},
... {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"},
... {"pqr":"sports"}]
>>> from collections import Counter
>>> counts = Counter(key for dic in d for key in dic.keys())
>>> counts
Counter({'pqr': 5, 'abc': 3, 'xyz': 1})
>>> for key in counts:
... print (key, counts[key])
...
xyz 1
abc 3
pqr 5
什么你所描述的 - 与每个键多个值的列表 - 可以由下面得到更好的可视化这样的：
{'abc': ['movies', 'sports', 'music'],
'xyz': ['music'],
'pqr': ['music', 'movies', 'sports', 'news']
}
在这种情况下，你必须做一些更多的工作要插入：
看到[]（空单）新的密钥if value in，看是否被检查的值存在于列表.append()它这也导致了简单的方法来统计存储的元素总数：
# Pseudo-code
for myKey in myDict.keys():
print "{0}: {1}".format(myKey, len(myDict[myKey])
使用collections.Counter。假设您有一个项目词典的列表...
from collections import Counter
listOfDictionaries = [{'abc':'movies'}, {'abc':'sports'}, {'abc':'music'},
{'xyz':'music'}, {'pqr':'music'}, {'pqr':'movies'},
{'pqr':'sports'}, {'pqr':'news'}, {'pqr':'sports'}]
Counter(list(dict)[0] for dict in zzz)
不需要使用计数器。您可以通过这种方式实现：
# input dictionary
d=[{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"}, {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"}, {"pqr":"sports"}]
# fetch keys
b=[j[0] for i in d for j in i.items()]
# print output
for k in list(set(b)):
print "{0}: {1}".format(k, b.count(k))
大厦@akashdeep解决方案，它采用了一套，但给出了一个错误的结果，因为没有在问题中提到的“明显”的要求计算（pqr应该是4，不是5 ）。
# dictionary
d=[{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"}, {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"}, {"pqr":"sports"}]
# merged dictionary
c = {}
for i in d:
for k,v in i.items():
try:
c[k].append(v)
except KeyError:
c[k] = [v]
# counting and printing
for k,v in c.items():
print "{0}: {1}".format(k, len(set(v)))
这会给出正确的：
xyz: 1
abc: 3
pqr: 4
你的意思是你有字典的名单？还是它复制不正确？ – thegrinner 2013-05-06 20:05:39
这不是一本字典，它至多是一个字典列表（它只包含一个键/值对） - 真的吗？这是什么样的数据结构？我猜它实际上是'[{“abc”：“电影”}，...，对吧？ – 2013-05-06 20:05:55
@TimPietzcker没错。对不起，代表性错误 – user1189851 2013-05-06 20:06:55