Question

我将继续进行编码练习，这使我返回了一个字典，其中的键是单词的长度，而值是单词本身。这是通过拆分文本（该文本是传递给get_word_len_dict（text）函数的参数）并计算字符数来完成的。然后将长度排序并输出到print_dict_in_key_order（a_dict）。

我得到这样的输出：

2 : ['to', 'is']
3 : ['why', 'you', 'say', 'are', 'but', 'the', 'wet']
4 : ['does', 'when', 'four', 'they', 'have']
5 : ['there', 'stars', 'check', 'paint']
7 : ['someone', 'believe', 'billion']

哪个看起来正确，但是如果我想按字母顺序对列表中的值进行排序怎么办？这意味着以大写字母开头的单词也应优先考虑。例如。 ['May'，'and']。

理想情况下，我希望这样的输出具有按字母顺序排列的值：

2 : ['is', 'to']
3 : ['are', 'but', 'say', 'the', 'wet', 'why', 'you']
4 : ['does', 'four', 'have', 'they', 'when']
5 : ['check', 'paint', 'stars', 'there']
7 : ['believe', 'billion', 'someone']

到目前为止，我已经设法在print_dict_in_key_order（a_dict）内对键进行了排序，但是如果我也想对值进行排序，则不确定如何处理？

def get_word_len_dict(text):
    dictionary = {}
    word_list = text.split()
    for word in word_list:
        letter = len(word)

        dictionary.setdefault(letter,[])

        if word not in dictionary[letter]:
            dictionary[letter].append(word)

    return dictionary

def test_get_word_len_dict():
    text = 'why does someone believe you when you say there are four billion stars but they have to check when you say the paint is wet'
    the_dict = get_word_len_dict(text)
    print_dict_in_key_order(the_dict)


def print_dict_in_key_order(a_dict): 
    all_keys = list(a_dict.keys()) 
    all_keys.sort() 
    for key in all_keys: 
        print(key, ":", a_dict[key])

Answer 1

鉴于此命令

d = {
    2: ['to', 'is'],
    3: ['why', 'you', 'say', 'are', 'but', 'the', 'wet'],
    4: ['does', 'when', 'four', 'they', 'have'],
    5: ['there', 'stars', 'check', 'paint'],
    7: ['someone', 'believe', 'billion'],
    }

您可以像这样对值进行排序：

{k: sorted(v) for k, v in d.items()}

输出（通过pprint）：

{2: ['is', 'to'],
 3: ['are', 'but', 'say', 'the', 'wet', 'why', 'you'],
 4: ['does', 'four', 'have', 'they', 'when'],
 5: ['check', 'paint', 'stars', 'there'],
 7: ['believe', 'billion', 'someone']}

尽管如果您只关心在打印时对其进行排序，则只需在代码中更改此行即可：

print(key, ":", a_dict[key])

对此：

print(key, ":", sorted(a_dict[key]))

Answer 2

您想要做的是按长度分组，然后按值排序（因为按字典顺序比较，大写字母比小写字母“小”），然后从每个组中删除重复项并将所有内容放入dict中理解力。

请注意，itertools.groupby与pandas中的类似功能不同，它将不连续的组视为不同的，因此我们需要先按长度排序。

示例：

from itertools import groupby
from pprint import pprint

def solution(sentence):
    sorted_words = sorted(sentence.split(' '), key=len)
    return {length: sorted(set(words)) for length, words in groupby(sorted_words, len)}

sentence =  'Why does someone believe you when you say there are four billion stars but they have to check when you say the paint is wet'

pprint(solution(sentence))

输出：

{2: ['is', 'to'],
 3: ['Why', 'are', 'but', 'say', 'the', 'wet', 'you'],
 4: ['does', 'four', 'have', 'they', 'when'],
 5: ['check', 'paint', 'stars', 'there'],
 7: ['believe', 'billion', 'someone']}

请注意，'Why'位于其他字母之前，因为它以大写字母开头，其余字母按字母顺序排序。

如果要保留功能结构，只需对字典中的每个list进行排序：

def get_word_len_dict(text):
    dictionary = {}
    word_list = text.split()
    for word in word_list:
        letter = len(word)

        dictionary.setdefault(letter,[])

        if word not in dictionary[letter]:
            dictionary[letter].append(word)

    for words in dictionary.values():
        words.sort()

    return dictionary

Answer 3

d = {
    2: ['to', 'is'],
    3: ['why', 'you', 'say', 'are', 'but', 'the', 'wet'],
    4: ['does', 'when', 'four', 'they', 'have'],
    5: ['there', 'stars', 'check', 'paint'],
    7: ['someone', 'believe', 'billion'],
    }

for i in d:
    d[i].sort()
print(d)

输出

   {
    2: ['is', 'to'],
    3: ['are', 'but', 'say', 'the', 'wet', 'why', 'you'],
    4: ['does', 'four', 'have', 'they', 'when'], 
    5: ['check', 'paint', 'stars', 'there'], 
    7: ['believe', 'billion', 'someone']
    }

如何对列表中的字典值进行排序？

3 个答案: