Python 将行拆分为段落

输入：行的列表

输出：行列表，是在（一个或多个）空行处拆分的输入列表

这是迄今为止我所拥有的最不丑陋的解决方案：

split_at_empty(lines):
    paragraphs = []
    p = []
    def flush():
        if p:
            paragraphs.append(p)
        p = []
    for l in lines:
        if l:
            p.append(l)
        else:
            flush()
    flush()
    return paragraphs

必须有更好的解决方案（甚至可能是功能性的）！有人吗

输入列表示例：

['','2','3','','5','6','7','8','','','11']

输出：

[['2','3'],['5','6','7','8'],['11']]

---

您可以将列表合并为字符串，然后重新拆分：

>>> a = ['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
>>> [x.strip().split(' ') for x in ' '.join(a).split('  ')]
[['2', '3'], ['5', '6', '7', '8'], ['11']]

您可能应该使用正则表达式来捕获任意数量的空格（我在'11'之前添加了另一个）：

---

您可以将列表合并为字符串，然后重新拆分：

>>> a = ['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
>>> [x.strip().split(' ') for x in ' '.join(a).split('  ')]
[['2', '3'], ['5', '6', '7', '8'], ['11']]

您可能应该使用正则表达式来捕获任意数量的空格（我在'11'之前添加了另一个）：

---

以下是基于生成器的解决方案：

def split_at_empty(lines):
   sep = [0] + [i for (i,l) in enumerate(lines) if not l] + [len(lines)]
   for start, end in zip(sep[:-1], sep[1:]):
      if start + 1 < end:
         yield lines[start+1:end]

它产生

['2', '3']
['5', '6', '7', '8']
['11']

---

以下是基于生成器的解决方案：

def split_at_empty(lines):
   sep = [0] + [i for (i,l) in enumerate(lines) if not l] + [len(lines)]
   for start, end in zip(sep[:-1], sep[1:]):
      if start + 1 < end:
         yield lines[start+1:end]

它产生

['2', '3']
['5', '6', '7', '8']
['11']

结果

['Princess Maria Amelia of Brazil (1831\x961853)']

['was the daughter of Dom Pedro I,', "founder of Brazil's independence and its first emperor,"]

['and Amelie of Leuchtenberg.']

["The only child from her father's second marriage,", 'Maria Amelia was born in France', "following Pedro I's 1831 abdication in favor of his son Dom Pedro II."]

['Before Maria Amelia was a month old, Pedro I left for Portugal', 'to restore its crown to his eldest daughter Dona Maria II.', "He defeated his brother Miguel I (who had usurped Maria II's throne),", 'only to die a few months later of tuberculosis.']

[['2', '3'], ['5', '6', '7', '8'], ['11']]

['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
[['2', '3'], ['5', '6', '7', '8'], ['11']]

['5055', '', '', '2', '54', '87', '', '1', '2', '5', '8', '', '']
[['5055'], ['2', '54', '87'], ['1', '2', '5', '8']]

['AAAAA', 'BB', '', 'HU', 'JU', 'GU']
[['AAAAA', 'BB'], ['HU', 'JU', 'GU']]

另一种方法是按列表行事：

li = [ '', '2', '3', '', '5', '6', '7', '8', '', '', '11']

lo = ['5055','','','2','54','87','','1','2','5','8','','']

lu = ['AAAAA','BB','','HU','JU','GU']

def selines(L):
    ye = []
    for x in L:
        if x:
            ye.append(x)
        elif ye:
            yield ye ; ye = []
    if ye:
        yield ye



for lx in (li,lo,lu):
    print lx
    print list(selines(lx))
    print

结果

['Princess Maria Amelia of Brazil (1831\x961853)']

['was the daughter of Dom Pedro I,', "founder of Brazil's independence and its first emperor,"]

['and Amelie of Leuchtenberg.']

["The only child from her father's second marriage,", 'Maria Amelia was born in France', "following Pedro I's 1831 abdication in favor of his son Dom Pedro II."]

['Before Maria Amelia was a month old, Pedro I left for Portugal', 'to restore its crown to his eldest daughter Dona Maria II.', "He defeated his brother Miguel I (who had usurped Maria II's throne),", 'only to die a few months later of tuberculosis.']

[['2', '3'], ['5', '6', '7', '8'], ['11']]

['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
[['2', '3'], ['5', '6', '7', '8'], ['11']]

['5055', '', '', '2', '54', '87', '', '1', '2', '5', '8', '', '']
[['5055'], ['2', '54', '87'], ['1', '2', '5', '8']]

['AAAAA', 'BB', '', 'HU', 'JU', 'GU']
[['AAAAA', 'BB'], ['HU', 'JU', 'GU']]

结果

['Princess Maria Amelia of Brazil (1831\x961853)']

['was the daughter of Dom Pedro I,', "founder of Brazil's independence and its first emperor,"]

['and Amelie of Leuchtenberg.']

["The only child from her father's second marriage,", 'Maria Amelia was born in France', "following Pedro I's 1831 abdication in favor of his son Dom Pedro II."]

['Before Maria Amelia was a month old, Pedro I left for Portugal', 'to restore its crown to his eldest daughter Dona Maria II.', "He defeated his brother Miguel I (who had usurped Maria II's throne),", 'only to die a few months later of tuberculosis.']

[['2', '3'], ['5', '6', '7', '8'], ['11']]

['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
[['2', '3'], ['5', '6', '7', '8'], ['11']]

['5055', '', '', '2', '54', '87', '', '1', '2', '5', '8', '', '']
[['5055'], ['2', '54', '87'], ['1', '2', '5', '8']]

['AAAAA', 'BB', '', 'HU', 'JU', 'GU']
[['AAAAA', 'BB'], ['HU', 'JU', 'GU']]

另一种方法是按列表行事：

li = [ '', '2', '3', '', '5', '6', '7', '8', '', '', '11']

lo = ['5055','','','2','54','87','','1','2','5','8','','']

lu = ['AAAAA','BB','','HU','JU','GU']

def selines(L):
    ye = []
    for x in L:
        if x:
            ye.append(x)
        elif ye:
            yield ye ; ye = []
    if ye:
        yield ye



for lx in (li,lo,lu):
    print lx
    print list(selines(lx))
    print

结果

['Princess Maria Amelia of Brazil (1831\x961853)']

['was the daughter of Dom Pedro I,', "founder of Brazil's independence and its first emperor,"]

['and Amelie of Leuchtenberg.']

["The only child from her father's second marriage,", 'Maria Amelia was born in France', "following Pedro I's 1831 abdication in favor of his son Dom Pedro II."]

['Before Maria Amelia was a month old, Pedro I left for Portugal', 'to restore its crown to his eldest daughter Dona Maria II.', "He defeated his brother Miguel I (who had usurped Maria II's throne),", 'only to die a few months later of tuberculosis.']

[['2', '3'], ['5', '6', '7', '8'], ['11']]

['', '2', '3', '', '5', '6', '7', '8', '', '', '11']
[['2', '3'], ['5', '6', '7', '8'], ['11']]

['5055', '', '', '2', '54', '87', '', '1', '2', '5', '8', '', '']
[['5055'], ['2', '54', '87'], ['1', '2', '5', '8']]

['AAAAA', 'BB', '', 'HU', 'JU', 'GU']
[['AAAAA', 'BB'], ['HU', 'JU', 'GU']]

---

比原版稍微丑一点：

def split_at_empty(lines):
    r = [[]]
    for l in lines:
        if l:
            r[-1].append(l)
        else:
            r.append([])
    return [l for l in r if l]

---

（最后一行去掉了原本要添加的空列表。）

比原来的略不难看：

def split_at_empty(lines):
    r = [[]]
    for l in lines:
        if l:
            r[-1].append(l)
        else:
            r.append([])
    return [l for l in r if l]

---

（最后一行删除了原本会添加的空列表。）

对于列表理解痴迷者

def split_at_empty(L):
    return [L[start:end+1] for start, end in zip(
        [n for n in xrange(len(L)) if L[n] and (n == 0 or not L[n-1])],
        [n for n in xrange(len(L)) if L[n] and (n+1 == len(L) or not L[n+1])]
        )]

或者更好

def split_at_empty(lines):
    L = [i for i, a in enumerate(lines) if not a]
    return [lines[s + 1:e] for s, e in zip([-1] + L, L + [len(lines)]) 
            if e > s + 1]

---

而对于那些痴迷于理解的人

def split_at_empty(L):
    return [L[start:end+1] for start, end in zip(
        [n for n in xrange(len(L)) if L[n] and (n == 0 or not L[n-1])],
        [n for n in xrange(len(L)) if L[n] and (n+1 == len(L) or not L[n+1])]
        )]

或者更好

def split_at_empty(lines):
    L = [i for i, a in enumerate(lines) if not a]
    return [lines[s + 1:e] for s, e in zip([-1] + L, L + [len(lines)]) 
            if e > s + 1]

---

---

发布输入列表的示例。@Jo因此您的“解决方案”不起作用：

flush（）

中的局部

p

负责

UnboundLocalError:赋值前引用的局部变量“p”

。那不是serious@eyquem. 我的错。太过沉迷于JavaScript了。要让它工作，我们必须让它更难看一点。@Jo好吧，你是一个很好的人张贴了你的输入列表样本。@Jo所以你的“解决方案”不起作用：local

p
in
flush（）
负责
UnboundLocalError:赋值前引用的局部变量“p
。那不是serious@eyquem. 我的错。太过沉迷于JavaScript了。要让它工作，我们必须让它更难看一点。@Jo好吧，你是个不错的人我想到过这个，但你不觉得它有点复杂，有太多的开销和长的线路，唯一的好处是线路少？我想到过这个，但是你不觉得它有点复杂，而且有太多的开销和很长的线路，唯一的好处是线路少？也许这是最好的一个！使用发电机可以让它更干净，谢谢。现在，这是一个被接受的答案。请把它编辑得简洁明了，你会得到回复：）@Jo你好，我仔细考虑了你的评论。我在两天前更正了上述代码。今天我还纠正了我的其他答案。因为你是对的，我倾向于写太长的答案。我甚至删除了我刚才在这里写的评论，没有兴趣用我这些无用的评论来干扰stackoverflow的记忆。谢谢你指出我的缺点，我会提醒你，也许这是最好的！使用发电机可以让它更干净，谢谢。现在，这是一个被接受的答案。请把它编辑得简洁明了，你会得到回复：）@Jo你好，我仔细考虑了你的评论。我在两天前更正了上述代码。今天我还纠正了我的其他答案。因为你是对的，我倾向于写太长的答案。我甚至删除了我刚才在这里写的评论，没有兴趣用我这些无用的评论来干扰stackoverflow的记忆。谢谢你指出了我的缺点，我会提醒你，还不错，真的！开销还可以。我比较喜欢这个简单。唯一的问题是如果输入列表很大。真的不错！开销还可以。我比较喜欢这个简单。唯一的问题是输入列表是否庞大。不幸的是，第一个列表是错误的。第二个很好，但不能很好地使用列表生成器。这两个都适用于原始示例输入（和其他输入）。您使用的是什么输入列表？对不起，您是对的，但是使用不同索引的
L
和
行有点混乱。而且，与
L[n]！=“相比，
L[n]更容易阅读，而不是L[n+1]
”而L[n+1]=''
（也许你想把我的更改排除在外）不幸的是，第一个错误。第二个很好，但不能很好地使用列表生成器。这两个都适用于原始示例输入（和其他输入）。您使用的是什么输入列表？对不起，您是对的，但是使用不同索引的
L
和
行有点混乱。而且，与
L[n]！=“相比，
L[n]更容易阅读，而不是L[n+1]
”和L[n+1]=''
（也许你想删除我的更改）