我有一個計算Python中每個鍵的不同值的問題。蟒蛇字典的唯一值計數
我有一個字典d像
[{"abc":"movies"}, {"abc": "sports"}, {"abc": "music"}, {"xyz": "music"}, {"pqr":"music"}, {"pqr":"movies"},{"pqr":"sports"}, {"pqr":"news"}, {"pqr":"sports"}]
我需要單獨打印每每個鍵的不同值的數目。
這意味着我將要打印
abc 3
xyz 1
pqr 4
請幫助。
謝謝
使用collections.Counter() instance，一些鏈接在一起：
from collections import Counter
from itertools import chain
counts = Counter(chain.from_iterable(e.keys() for e in d))
這保證了在你的輸入列表中有多個鍵的字典進行正確計數。
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})
或與輸入的詞典多個鍵：
>>> 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，該目錄排序的元件，反向其計數.most_common() method訂購：
for key, count in counts.most_common():
print '{}: {}'.format(key, count)
# prints
# 5: pqr
# 3: abc
# 1: xyz
請注意，計數器]（http://docs.python.org/2/library/collections.html#collections.Counter）類是在Python 2.7中引入的。有[backport]（http://code.activestate.com/recipes/576611-counter-class/）。我想你[知道這件事]（Martijn）（http://stackoverflow.com/a/13311111/566644）。 – 2013-05-06 20:14:50
@ LauritzV.Thaulow：以其他方式作爲[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
什麼你所描述的 - 與每個鍵多個值的列表 - 可以由下面得到更好的可視化這樣的：
{'abc': ['movies', 'sports', 'music'],
'xyz': ['music'],
'pqr': ['music', 'movies', 'sports', 'news']
}
在這種情況下，你必須做一些更多的工作要插入：
看到[]（空單）新的密鑰if value in，看是否被檢查的值存在於列表.append()它這也導致了簡單的方法來統計存儲的元素總數：
# Pseudo-code
for myKey in myDict.keys():
print "{0}: {1}".format(myKey, len(myDict[myKey])
使用collections.Counter。假設您有一個項目詞典的列表...
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)
不需要使用計數器。您可以通過這種方式實現：
# 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))
大廈@akashdeep解決方案，它採用了一套，但給出了一個錯誤的結果，因爲沒有在問題中提到的「明顯」的要求計算（pqr應該是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)))
這會給出正確的：
xyz: 1
abc: 3
pqr: 4
你的意思是你有字典的名單？還是它複製不正確？ – thegrinner 2013-05-06 20:05:39
這不是一本字典，它至多是一個字典列表（它只包含一個鍵/值對） - 真的嗎？這是什麼樣的數據結構？我猜它實際上是'[{「abc」：「電影」}，...，對吧？ – 2013-05-06 20:05:55
@TimPietzcker沒錯。對不起，代表性錯誤 – user1189851 2013-05-06 20:06:55