Question

我试图使用列表推导将嵌套列表划分为两个嵌套列表。如果不将内部列表转换为字符串，我无法这样做，这反过来又破坏了我以后访问/打印/控制值的能力。

我试过这个::

paragraphs3 = [['Page: 2', 'Bib: Something', 'Derived:  This n that'], ['Page: 3', 'Bib: Something', 'Argument: Wouldn't you like to know?'], ...]

derived = [k for k in paragraphs3 if 'Derived:' in k]
therest = [k for k in paragraphs3 if 'Derived:' not in k]

会发生的是，整个paragraph3 = []最终会出现在where = []中，除非我这样做：

for i in paragraphs3:
    i = str(i)
    paragraphs4.append(i)

如果我然后将paragraph4提供给列表理解，我会得到两个列表，就像我想要的那样。但是它们不再是嵌套列表了：

    for i in therest:
        g.write('\n'.join(i))
        g.write('\n\n')

写每个！角色！ inst = []在一个单独的行中：

'
P
a
g
e
:

2
'

因此，我正在寻找一种更好的方法来分割段落3 ......或者解决方案可能在其他地方？我要找的最终结果/输出是：

Page: 2
Bib: Something
Derived: This n that

Page: 3
Bib: Something
.
.
.

Answer 1

此代码根据子列表是否包含以'Derived:'开头的字符串来分隔子列表。

paragraphs3 = [['Page: 2', 'Bib: Something', 'Derived:  This n that'], ['Page: 3', 'Bib: Something', "Argument: Wouldn't you like to know?"], ]

def show(paragraphs):
    for para in paragraphs:
        print('\n'.join(para), '\n')

derived = []
therest = []

print('---input---')
show(paragraphs3)

for para in paragraphs3:
    if any(item.startswith('Derived:') for item in para):
        derived.append(para)
    else:
        therest.append(para)

print('---derived---')
show(derived)

print('---therest---')
show(therest)

输出

---input---
Page: 2
Bib: Something
Derived:  This n that 

Page: 3
Bib: Something
Argument: Wouldn't you like to know? 

---derived---
Page: 2
Bib: Something
Derived:  This n that 

---therest---
Page: 3
Bib: Something
Argument: Wouldn't you like to know?

此代码最重要的部分是

`any(item.startswith('Derived:') for item in para)`

这会迭代para（当前段落）中的各个字符串，并在找到以True开头的字符串后立即返回'Derived:'。

FWIW，for循环可以缩减为：

for para in paragraphs3:
    (therest, derived)[any(item.startswith('Derived:') for item in para)].append(para)

因为False和True分别计算为0和1，所以它们可用于索引(therest, derived)元组。然而，许多人会认为这是不可读的。 :)

Answer 2

您编写的代码几乎是正确的。您需要检查列表的第3个元素中是否存在'Derived:'。 k基本上包含paragraphs3

的第一个元素

>>> paragraphs3 = [['Page: 2', 'Bib: Something', 'Derived:  This n that'], ['Page: 3', 'Bib: Something', 'Argument: Wouldn\'t you like to know?']]
>>> paragraphs3[0]
['Page: 2', 'Bib: Something', 'Derived:  This n that']
>>> paragraphs3[0][2] # Here is where you want to check the condition
'Derived:  This n that'

所以你要做的就是将条件改为if 'Derived:' in k[2]。

>>> [k for k in paragraphs3 if 'Derived:' in k[2]]
[['Page: 2', 'Bib: Something', 'Derived:  This n that']]

>>> [k for k in paragraphs3 if 'Derived:' not in k[2]]
[['Page: 3', 'Bib: Something', "Argument: Wouldn't you like to know?"]]

Answer 3

解决方案

'Dim i As Long

'For i = 5 To 25

'If Not IsEmpty(Sheets("RMDA").Range("D" & i)) Then _

Sheets("Overview").Range("D" & i) = Sheets("RMDA").Range("D" & i)

'Next i

详细说明

复制整个列表：

derived = [l for l in paragraphs3 if any(filter(lambda k: 'Derived: ' in k, l))]
therest = [l for l in paragraphs3 if any(filter(lambda k: 'Derived: ' not in k, l))]

使用条件复制列表：

[l for l in paragraph3]

功能[l for l in paragraph3 if sublist_contains('Derived: ', l)]尚未实现，所以让我们实现它。

仅检索与sublist_contains匹配的项目：

condition_check

由于filter(condition_check, l)可以表示为lambda函数：

condition_check

将结果转换为布尔值（如果找到至少一个匹配条件的项目，则为True）：

filter(lambda k: 'Derived: ' in k, l)

用结果内联代码替换any(filter(lambda k: 'Derived: ' in k, l))：

sublist_contains

Answer 4

在我看来，这是最直接的做法：

[p for p in paragraphs3 if 'Derived:' in '\n'.join(p)]
[p for p in paragraphs3 if 'Derived:' not in '\n'.join(p)]

然而，如果你愿意的话，你可以获得更多的爱好者，并将其拉到一行（虽然它会比必要的更复杂）。

result = {k:[p for p in paragraphs3 if ('Derived:' in '\n'.join(p)) == test]  for k,test in {'derived': True, 'therest': False}.items()}

这会产生dict 'derived'和'therest'作为键。现在你可以这样做：

for k,p in result.items():
    print(k)
    for i in p:
        print(''.join(i))

Answer 5

看起来你的内心清单有结构;列表本身是一个值，而不仅仅是一个不相关的值列表。考虑到这一点，您可以编写一个类来表示该数据。

paragraphs3 = [['Page: 2', 'Bib: Something', 'Derived:  This n that'], ['Page: 3', 'Bib: Something', 'Argument: Wouldn\'t you like to know?'], ...]

class Paragraph(object):
    def __init__(self, page, bib, extra):
        self.page = page
        self.bib = bib
        self.extra = extra

    @property
    def is_derived(self):
        return 'Derived: ' in self.extra

paras = [Paragraph(p) for p in paragraphs3]

然后，您可以使用itertools中的partition配方将该列表拆分为两个迭代器。

def partition(pred, iterable):
    'Use a predicate to partition entries into false entries and true entries'
    # partition(is_odd, range(10)) --> 0 2 4 6 8   and  1 3 5 7 9
    t1, t2 = tee(iterable)
    return filterfalse(pred, t1), filter(pred, t2)

(not_derived_paras, derived_paras) = partition(lambda p: p.is_derived, paras)

在python中包含字符串的列表中提取列表

5 个答案:

输出

解决方案

详细说明