如何将x个长度为x的单词添加到不同的嵌套列表中

Question

我有一项工作，需要根据单词的长度将单词添加到相应的列表中。也就是说，所有长度为1的单词将进入列表1，长度为2的单词将进入列表2，依此类推……

下面是我目前拥有的代码。如你所见，我已经创建了一个带有L个空存储桶的列表，其思想是让每个长度的单词都有对应的存储桶。这就是我被卡住的地方。如果不知道将有多少个存储桶，我不知道如何添加它们。我对Python非常陌生，任何帮助都将不胜感激！

def empty_buckets(n): 
    """Return a list with n empty lists. Assume n is a positive integer. """ 
    buckets = [] 
    for bucket in range(n): 
        buckets.append([]) 
    return buckets 

计算所有单词的最大长度L。

longest = ''
    for L in words:
        if len(L) > len(longest):
            longest = L
    return longest

创建一个包含L个空列表(存储桶)的列表。

buckets = empty_buckets(L)

Answer 1

您可以使用max()并提供len的键函数来获取单词列表中最长的单词。

您可以为“空”单词再创建一个存储桶，并使用for循环将所有单词排序到存储桶中，并使用len(word)将所有单词索引到存储桶中

# create some demo strings and add some other words
words = [ str(10**k) for k in range(10)]
words.extend(["this","should","work","out","somehow"])

print(words)  # ['1', '10', '100', '1000', '10000', '100000', '1000000', '10000000',
              #  '100000000', '1000000000', 'this', 'should', 'work', 'out', 'somehow']

longest = len(max(words,key=len)) # get the length of the longest word

# create a empty bucket for "" and one bucket for length 1 up to longest
bins = [None] + [ [] for _ in range(longest+1)]  

# loop over words and put then in the bin at index len(word)
for w in words:
    bins[len(w)].append(w)

print(bins)

输出：

[None, ['1'], ['10'], ['100', 'out'], ['1000', 'this', 'work'], ['10000'], 
       ['100000', 'should'], ['1000000', 'somehow'], ['10000000'], 
       ['100000000'], ['1000000000']]

Doku：

• max(iterable, key=len)
• len()
• range()

Answer 2

buckets = [0] * longest # this will make a list of longest size

然后，在每个元素中创建一个列表，我使用列表的第一个元素来记录该存储桶的计数。

for i in range(longest): buckets[i] = [0]

然后，您需要将单词添加到桶中。

for L in words: buckets[len(L)][0] += 1 # increasing the count of that bucket buckets[len(L)].append(L) # Adding the word to that bucket

下面是一个示例：

longest = 10
words = ['this', 'that', 'foremost']
buckets = [0] * longest # this will make a list of longest size 
for i in range(longest):
   buckets[i] = [0]
for L in words:
   buckets[len(L)][0] += 1 # increasing the count of that bucket
   buckets[len(L)].append(L) # Adding the word to that bucket 

要访问任何计数，只需使用buckets[number][0]，而要访问所有单词，可以从buckets[number][1]开始循环该计数。

Answer 3

正如我之前在评论中提到的，我使用字典来解决这个问题。

在这里，您不需要使用任何外部函数创建空列表，因为我们不知道实际长度。

所以你可以尝试这样做。

您可以访问https://rextester.com/ZQKA28350在线运行代码。

def add_words_to_bucket(words): 
    d = {}

    for word in words: 
        l = len(word)
        if l in d: 
            d[l].append(word) 
        else: 
            i = 0
            while l >= 0 and not l in d:
                if not i: 
                    d[l] = [word]
                else: 
                    d[l] = []
                l = l - 1
                i += 1
    return d

def get_as_list(d): 
    bucket = [d[i] for i in range(0, len(d))]
    return bucket


words = ["a",  "git", "go", "py", "java", "paper", "ruby", "r"]
d = add_words_to_bucket(words) 
bucket = get_as_list(d)
print(d) # {0: [], 1: ['a', 'r'], 2: ['go', 'py'], 3: ['git'], 4: ['java', 'ruby'], 5: ['paper']}
print(bucket) # [[], ['a', 'r'], ['go', 'py'], ['git'], ['java', 'ruby'], ['paper']]


words2 = ["a",  "git", "go", "py", "", "java", "paper", "ruby", "r","TheIpMan", ""]
d2 = add_words_to_bucket(words2)
bucket2 = get_as_list(d2)
print(d2) # {0: ['', ''], 1: ['a', 'r'], 2: ['go', 'py'], 3: ['git'], 4: ['java', 'ruby'], 5: ['paper'], 6: [], 7: [], 8: ['TheIpMan']}
print(bucket2) # [['', ''], ['a', 'r'], ['go', 'py'], ['git'], ['java', 'ruby'], ['paper'], [], [], ['TheIpMan']]