如何将列表中的元素组合到新的嵌套列表中？

Question

我有一个这样的列表“r”：

[["", 1], ["this is a text line", 2], ["this is a text line", 3], ["this is a text line", 4], ["", 5], ["", 6], ["this is a text line", 7],["this is a text line", 8], ["this is a text line", 9], ["this is a text line", 10], ["", 11], ["this is a text line", 12], ["this is a text line", 13], ["this is a text line", 14], ["", 15], ["this is a text line", 16], ["this is a text line", 17], ["this is a text line", 18], ["", 19]]

要知道我的空行和带有文本的行在哪里，我过滤了我的列表：

empty = [x[1] for x in r if regex.search("^\s*$", x[0])]
text = [x[1] for x in r if regex.search("\S", x[0])]

输出：

empty = [1, 5, 6, 11, 15, 19]
text= [2, 3, 4, 7, 8, 9, 10, 12, 13, 14, 16, 17, 18]

我想要做的是组合文本中的数字，如果它们是按顺序排列的 (text[i]-text[i+1]) = +1（为了定义段落）：

finaltext = [[2, 3, 4], [7, 8, 9, 10], [12, 13, 14], [16, 17, 18]]
finaltext including empty = [[2, 3, 4, 5, 6], [7, 8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]]

如何根据条件对列表中的元素进行分组？

Answer 1

使用itertools.groupby

from itertools import groupby, zip_longest
grp_list = [list(g) for k,g in groupby(r, lambda x:x[0]=='')]
grp_list = grp_list[1:] if r[0][0] == '' else grp_list
text = [[j[1] for j in i] for i in grp_list]

finaltext = text[::2]
print (finaltext)
#[[2, 3, 4], [7, 8, 9, 10], [12, 13, 14], [16, 17, 18]]

finaltext_including_empty = [i+j for i,j in zip_longest(text[::2], text[1::2], fillvalue=[])]
print (finaltext_including_empty)
#[[2, 3, 4, 5, 6], [7, 8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]]

groupby 根据lambda x:x[0]=='' 的条件将列表分组为子列表块，这意味着创建一个列表块直到您看到一个空字符串为止，并遵循此规则直到结束，如下所示

[[['', 1]], [['this is a text line', 2], ['this is a text line', 3], ['this is a text line', 4]], [['', 5], ['', 6]],........]

Answer 2

没有任何modules的纯Python解决方案：

这可以使用modules 来完成，例如numpy 和groupby，但我认为没有它们会被调用，只需使用简单的Python。这是我的解决方案：

text = [2, 3, 4, 7, 8, 9, 10, 12, 13, 14, 16, 17, 18]
s = 0
finaltext = []
for i in range(len(text)-1):
    if text[i] + 1 != text[i+1]:
        finaltext.append(text[s:i+1])
        s = i+1

finaltext.append(text[s:])

将finaltext 设为：

[[2, 3, 4], [7, 8, 9, 10], [12, 13, 14], [16, 17, 18]]

---

更新

要同时获得lists（不知道为什么要这样做），您可以使用以下命令：

empty = [1, 5, 6, 11, 15, 19]
text = [2, 3, 4, 7, 8, 9, 10, 12, 13, 14, 16, 17, 18]
s = 0
finaltext = []
finaltext_including_empty = []
for i in range(len(text)-1):
    if text[i] + 1 != text[i+1]:
        finaltext.append(text[s:i+1])
        finaltext_including_empty.append(list(range(text[s], text[i+1])))
        s = i+1

finaltext.append(text[s:])
finaltext_including_empty.append(list(range(text[s],max(empty[-1]+1, text[-1]+1))))

这使得finaltext 与以前相同，finaltext_including_empty 为：

[[2, 3, 4, 5, 6], [7, 8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]]

Answer 3

pip install more_itertools

from more_itertools import chunked

empty = [1, 5, 6, 11, 15, 19]
text= [2, 3, 4, 7, 8, 9, 10, 12, 13, 14, 16, 17, 18]

finaltext_ = sorted(empty + text)

list(chunked(finaltext_,4))
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19]]