如何創建字典從另一個字典如果某些條件滿足

Question

從詞典：

{0: (u'Donald', u'PERSON'), 1: (u'John', u'PERSON'), 2: (u'Trump', u'PERSON'), 14: (u'Barack', u'PERSON'), 15: (u'Obama', u'PERSON'), 17: (u'Michelle', u'PERSON'), 18: (u'Obama', u'PERSON'), 30: (u'Donald', u'PERSON'), 31: (u'Jonh', u'PERSON'), 32: (u'Trump', u'PERSON')} 

我想創建另一個解釋如下：

{u'Donald John Trump': 2, u'Barack Obama':1, u'Michele Obama':1} 

這裏0,1,2和30 ，31,32個鍵正在增加1併發生兩次。每個發生14,15,17,18次。有什麼方法可以創建這樣的字典嗎？

Answer 1

我認爲你需要解決的主要問題是通過按照你描述的那樣通過對錶示增加的int序列的鍵進行分組來識別人員。

幸運的是，Python對此有a recipe。

from itertools import groupby 
from operator import itemgetter 
from collections import defaultdict 

dct = { 
    0: ('Donald', 'PERSON'), 
    1: ('John', 'PERSON'), 
    2: ('Trump', 'PERSON'), 
    14: ('Barack', 'PERSON'), 
    15: ('Obama', 'PERSON'), 
    17: ('Michelle', 'PERSON'), 
    18: ('Obama', 'PERSON'), 
    30: ('Donald', 'PERSON'), 
    31: ('John', 'PERSON'), 
    32: ('Trump', 'PERSON') 
} 

persons = defaultdict(int) # Used for conveniance 
keys = sorted(dct.keys()) # So groupby() can recognize sequences 

for k, g in groupby(enumerate(keys), lambda d: d[0] - d[1]): 
    ids = map(itemgetter(1), g)    # [0, 1, 2], [14, 15], etc. 
    person = ' '.join(dct[i][0] for i in ids) # "Donald John Trump", "Barack Obama", etc 
    persons[person] += 1 

print(persons) 
# defaultdict(<class 'int'>, 
#  {'Barack Obama': 1, 
#   'Donald John Trump': 2, 
#   'Michelle Obama': 1}) 

Answer 2

def add_name(d, consecutive_keys, result): 
    result_key = ' '.join(d[k][0] for k in consecutive_keys) 
    if result_key in result: 
     result[result_key] += 1 
    else: 
     result[result_key] = 1 

d = {0: (u'Donald', u'PERSON'), 1: (u'John', u'PERSON'), 2: (u'Trump', u'PERSON'), 
    14: (u'Barack', u'PERSON'), 15: (u'Obama', u'PERSON'), 
    17: (u'Michelle', u'PERSON'), 18: (u'Obama', u'PERSON'), 
    30: (u'Donald', u'PERSON'), 31: (u'John', u'PERSON'), 32: (u'Trump', u'PERSON')} 

sorted_keys = sorted(d.keys()) 
last_key = sorted_keys[0] 
consecutive_keys = [last_key] 
result = {} 
for i in sorted_keys[1:]: 
    if i == last_key + 1: 
     consecutive_keys.append(i) 
    else: 
     add_name(d, consecutive_keys, result) 
     consecutive_keys = [i]   
    last_key = i 
add_name(d, consecutive_keys, result) 

print(result) 

輸出

{'Donald John Trump': 2, 'Barack Obama': 1, 'Michelle Obama': 1}