如何處理在Python字典找不到鑰匙

Question

我有以下代碼：

from math import sqrt 
from collections import Counter 

def forSearch(): 
    words = {'bit':{1:3,2:4,3:19,4:0},'shoe':{1:0,2:0,3:0,4:0},'dog':{1:3,2:0,3:4,4:5}, 'red':{1:0,2:0,3:15,4:0}} 
    search = {'bit':1,'dog':3,'shoe':5} 
num_files = 4 

    file_relevancy = Counter() 
    c = sqrt(sum([x**2 for x in search.values()])) 
    for i in range(1, num_files+1): 
     words_ith_val = [words[x][i] for x in search.keys() ] 
     a = sum([search[key] * words[key][i] for key in search.keys()]) 
     b = sqrt(sum([x**2 for x in words_ith_val])) 
     file_relevancy[i] = (a/(b * c)) 

    return [x[0] for x in file_relevancy.most_common(num_files)] 

print forSearch() 

然而，這其中包含在搜索，但無法用語言文字方面的問題：

我想在這裏這樣說：

for i in range(1, num_files+1): 
    if corresponding key in words cannot be found 
     insert it and make its value = 0 
    words_ith_val = [words[x][i] for x in search.keys() ] 

那麼它應該工作？

除非其他人有更好的建議嗎？

Answer 1

collections.defaultdict

import collections 

D = collections.defaultdict(int) 
D['foo'] = 42 
print D['foo'], D['bar'] 

Answer 2

這個怎麼樣代碼：

if key not in words: 
    words[key] = {k+1: 0 for k in range(num_files)} 

在你的代碼，你可以嘗試做

for key in search.keys(): 
    if key not in words: 
     words[key] = {k+1: 0 for k in range(num_files)} 
    words_ith_val = [words[key][k + 1] for k in range(num_files)] 

Answer 3

可以使用defaultdict：

from collections import defaultdict 
d = defaultdict(int) 

這將初始化密鑰是在訪問和默認值創建的字典爲0，可以使用其他類型還有：

defaultdict(dict) 
defaultdict(list) 

他們將一個空的字典/列表進行初始化。 您也可以使用工廠方法覆蓋默認值。詳情請參閱https://docs.python.org/2/library/collections.html#collections.defaultdict。