Question

如何将每个单词的第一个字母保存为带有相关单词的字典中的密钥？

list = ['pine', 'dinner', 'liver', 'love', 'pick']

输出：

dictionary = {'p' : ['pine', 'pick'], 'd' : ['dinner'], 'l' : ['love', 'liver']}

Answer 1

使用默认Dict我们可以做到：

from collections import defaultdict

list_ = ['pine', 'dinner', 'liver', 'love', 'pick']

x = defaultdict(list)

for item in list_:
    x[item[0]].append(item)

print(x) 
# defaultdict(<class 'list'>, {'p': ['pine', 'pick'], 'd': ['dinner'], 'l': ['liver', 'love']})

然后，您可以像字典一样使用x：

print(x['p'])
#['pine', 'pick']

Answer 2

Dict = dict()

# iterate over the collection
for word in words:
    # get the first letter
    letter = word[0]

    # the default value for key 'letter' 
    # will be an empty list 
    # if the key isn't present yet
    # otherwise, nothing's changed
    Dict.setdefault(letter, [])

    # now you are sure that there's a list at that key
    Dict[letter].append(word)

Answer 3

我认为应该这样做。

dictionary = {}
list = ['pine', 'dinner', 'liver', 'love', 'pick']
for i in list:
    if i[0] not in dictionary.keys():
        dictionary[i[0]] = []
    dictionary[i[0]].append(i)

Answer 4

尝试：

list = ['pine', 'dinner', 'liver', 'love', 'pick']
d= dict()
for item1 in list:
    li=[]
    for item2 in list:
        if item1[0]==item2[0]:
            li.append(item2)
    d[item1[0]]= li
print d

将单词的第一个字母保存为键，将关联的单词保存为值？

4 个答案: