如何将列表切成不同长度的子列表

Question

上下文：我想制作多词搜索网格

我有一个这样的文本行输入（包含数字和字母字符串）：

2           <--- tells how many grids/boxes 
3           <--- tells rows/length
4           <--- tells columns/width 
catt        <--\
aata        <--- letters that will fill the box
tatc        <--/
cat         <--- the word that will be search inside the box
5           <--- and repeat but start with the rows
5
gogog
ooooo
godog
ooooo
gogog
dog

所有这些都在一个列表中作为输入

但是，我需要在里面传递变量。所以我假设我需要将列表拆分/切片为另一个列表，其中包含我需要的所有变量。

我想我需要像这样拆分 cat 和 dog 的变量：

#variables for cat and dog
rows, cols = [3, 5], [4, 5] #the numbers have to be integer
matrix = [['catt', 'aata', 'tatc'], ['gogog', 'ooooo', 'godog', 'ooooo', 'gogog']]
word = ['cat', 'dog']

这些是我需要的变量。但我不知道如何从上面的输入中拆分它。

如果有任何不同的方式，请随时解释。谢谢

Answer 1

我已将您的输入保存在一个名为“input.txt”的文件中，并使用了以下方法：

with open("input.txt", "r") as f:
    lines = [x.strip() for x in f.readlines()]

n_animals, other_lines = int(lines[0]), lines[1:]

rows, cols, matrix, word = [[] for _ in range(4)] # comment by @Stef - well spotted

while len(other_lines) > 0:
    rows.append(int(other_lines[0]))
    cols.append(int(other_lines[1]))
    matrix.append(list(map(lambda x: x[:cols[-1]], other_lines[2:2 + rows[-1]])))
    word.append(other_lines[2 + rows[-1]])
    other_lines = other_lines[2 + rows[-1] + 1:]
    if len(matrix) == n_animals:
        pass # Do we need to take any action here? like break?

print(rows)
print(cols)
print(matrix)
print(word)

输出

[3, 5]
[4, 5]
[['catt', 'aata', 'tatc'], ['gogog', 'ooooo', 'godog', 'ooooo', 'gogog']]
['cat', 'dog']

我的假设是你想用那个 width 变量做点什么，因此我把每个单词都剪成了cols[-1] 个字符。现在，你需要决定如果len(matrix) > n_animals 怎么办。

跟进

结合一些反馈以提高效率：

i = 0
while i < len(other_lines):
    rows.append(int(other_lines[i]))
    cols.append(int(other_lines[i + 1]))
    matrix.append(list(map(lambda x: x[:cols[-1]], other_lines[i + 2 : i + 2 + rows[-1]])))
    word.append(other_lines[i + 2 + rows[-1]])
    i += 2 + rows[-1] + 1
    if len(matrix) == n_animals:
        pass # Do we need to take any action here? like break?

Answer 2

使用next 遍历文件对象：

with open('input.txt') as f:
    n_animal = int(next(f).strip())
    rows, cols, matrices, words = [], [], [], []
    for _ in range(n_animals):
        n_row = int(next(f).strip())
        n_col = int(next(f).strip())
        rows.append(n_row)
        cols.append(n_col)
        matrix = [list(next(f).strip()) for _ in range(n_row)]
        matrices.append(matrix)
        words.append(next(f).strip())

print('rows, cols = ', rows, cols)
print('matrices = ', matrices)
print('words = ', words)

# rows, cols =  [3, 5] [4, 5]
# matrices =  [[['c', 'a', 't', 't'], ['a', 'a', 't', 'a'], ['t', 'a', 't', 'c']], [['g', 'o', 'g', 'o', 'g'], ['o', 'o', 'o', 'o', 'o'], ['g', 'o', 'd', 'o', 'g'], ['o', 'o', 'o', 'o', 'o'], ['g', 'o', 'g', 'o', 'g']]]
# words =  ['cat', 'dog']

注意：如果可以使用字符串列表而不是列表列表，则可以替换 matrix = 行：

# list of lists
matrix = [list(next(f).strip()) for _ in range(n_row)]

# list of strings
matrix = [next(f).strip() for _ in range(n_row)]

如果您的输入已经存储为字符串列表，而不是要读取的文件，您仍然可以在迭代器上使用 next：

lines = ['2', '3', '4', 'catt', ...]

f = iter(lines)

n_animal = int(next(f).strip())
rows, cols, matrices, words = [], [], [], []
for _ in range(n_animals):
    n_row = int(next(f).strip())
    n_col = int(next(f).strip())
    rows.append(n_row)
    cols.append(n_col)
    matrix = [list(next(f).strip()) for _ in range(n_row)]
    matrices.append(matrix)
    words.append(next(f).strip())

Answer 3

如果每行信息的位置始终是“固定的”，那么最简单的选择是将行转换为列表，然后具体引用每一行。比如：

data = text.splitlines()
grids = data[0]
rows = data[1]
cols = data[2]
letters = data[3:7]
repeat = data[7:9]
remain = data[9:]

print(grids, rows, cols, letters, repeat, remain)