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[英]Dictionary and function gene mapping output not returning expected frequencies
[英]Returning frequencies of words in a dictionary
我想我對如何解決這個函數有正確的想法,但我不確定為什么我沒有得到文檔字符串中顯示的所需結果。 有人可以幫我解決這個問題嗎?
def list_to_dict(word_list):
'''(list of str) -> dict
Given a list of str (a list of English words) return a dictionary that keeps
track of word length and frequency of length. Each key is a word length and
the corresponding value is the number of words in the given list of that
length.
>>> d = list_to_dict(['This', 'is', 'some', 'text'])
>>> d == {2:1, 4:3}
True
>>> d = list_to_dict(['A', 'little', 'sentence', 'to', 'create', 'a',
'dictionary'])
>>> d == {1:2, 6:2, 8:1, 2:1, 10:1}
True
'''
d = {}
count = 0
for i in range(len(word_list)):
length = len(i)
if length not in d:
count = count + length
d[length] = {count}
count += 1
return d
使用Counter
肯定是最好的選擇:
In [ ]: from collections import Counter
...: d = Counter(map(len, s))
...: d == {1:2, 6:2, 8:1, 2:1, 10:1}
Out[ ]: True
在不使用“花哨的東西”的情況下,我們使用生成器表達式,我認為它同樣有點奇特:
Counter(len(i) for i in s)
如果“通常”意味着使用for循環,我們可以這樣做:
d = {}
for i in s:
if len(i) not in d:
d[len(i)] = 1
else:
d[len(i)] += 1
只需對列表中的任何單詞執行循環。 在每次迭代中,如果長度不在dict中作為鍵,則創建值為1
新鍵,否則增加鍵的先前值:
def list_to_dict(word_list):
d = dict()
for any_word in word_list:
length = len(any_word)
if length not in d:
d[length] = 1
else:
d[length] += 1
return d
您可以使用字典理解來迭代s
,現在包含其原始元素的長度:
s = ['A', 'little', 'sentence', 'to', 'create', 'a', 'dictionary']
final_s = {i:len([b for b in s if len(b) == i]) for i in map(len, s)}
輸出:
{1: 2, 6: 2, 8: 1, 2: 1, 10: 1}
只是:
new_d = {}
for i in s:
new_d[len(i)] = 0
for i in s:
new_d[len(i)] += 1
輸出:
{8: 1, 1: 2, 2: 1, 10: 1, 6: 2}
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