Question

我有一个单词列表和一个整数'n'列表。如何将单词列表分成n个大小的块（不均匀）？

例如

words = ['apple', 'orange', 'oof', 'banana', 'apple', 'cherries', 'tomato']
n = ['2', '1', '4']

输出：

[['apple', 'orange'], ['oof'], ['banana', 'apple', 'cherries', 'tomato']]

Answer 1

另一个答案：

output = []
count = 0
for i in range(len(n)):
    chunk_size = int(n[i])
    chunk = []
    for j in range(chunk_size):
        index = (i * j) + j
        chunk.append(words[count])
        count = count + 1
    output.append(chunk)

print(output)

Answer 2

您可以对iter和next使用列表推导：

words = ['apple', 'orange', 'oof', 'banana', 'apple', 'cherries', 'tomato']
n = ['2', '1', '4']
new_words = iter(words)
result = [[next(new_words) for _ in range(int(i))] for i in n]

输出：

[['apple', 'orange'], ['oof'], ['banana', 'apple', 'cherries', 'tomato']]

Answer 3

简单的O(n)策略：

words = ['apple', 'orange', 'oof', 'banana', 'apple', 'cherries', 'tomato']
n = ['2', '1', '4']
start = 0
out = []
for num in n:
    num = int(num)
    out.append(words[start:start+num])
    start += num

Answer 4

如果您愿意导入numpy或itertools，则可以使用累积总和创建一个新列表，并将其用作索引以创建所需的输出：

# itertools.accumulate
from itertools import accumulate
m = [0] + list(accumulate([0] + n))

# numpy.cumsum
import numpy as np
m = np.cumsum([0] + n)

# Create new list
[words[i:j] for i, j in zip(m[:-1], m[1:])]

如何将列表拆分为n个大小的块，其中n是整数的可迭代列表？

4 个答案: