Question

我有以下代码：

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()

但是，这对搜索中包含的单词有疑问，而不是单词：

我想在这里说出这样的话：

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() ]

那它应该有用吗？

除非其他人有更好的建议吗？

Answer 1

collections.defaultdict

import collections

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

Answer 2

您可以使用defaultdict：

from collections import defaultdict
d = defaultdict(int)

这将初始化一个字典，其中密钥是在访问时创建的，默认值为0.您也可以使用其他类型：

defaultdict(dict)
defaultdict(list)

它们将使用空字典/列表进行初始化。您还可以使用工厂方法覆盖默认值。有关详细信息，请参阅https://docs.python.org/2/library/collections.html#collections.defaultdict。

Answer 3

这段代码怎么样：

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

在您的代码中，您可以尝试

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)]

如何处理Python字典中找不到的键

3 个答案: