在python中包含字符串的列表中提取列表

Question

我正在嘗試使用列表推導將嵌套列表划分為兩個嵌套列表。 如果不將內部列表轉換為字符串，我無法這樣做，這反過來又破壞了我以后訪問/打印/控制值的能力。

我試過這個::

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]

會發生的是整個paragraph3 = []最終在where = []，除非我做這樣的事情：

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

如果我然后將paragraph4提供給列表理解，我會得到兩個列表，就像我想要的那樣。 但是它們不再是嵌套列表了：

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

寫每個！角色！ inst = []在一個單獨的行中：

'
P
a
g
e
:

2
'

因此，我正在尋找一種更好的方法來分割段落3 ......或者解決方案可能在其他地方？ 我正在尋找的最終結果/輸出是：

Page: 2
Bib: Something
Derived: This n that

Page: 3
Bib: Something
.
.
.

Answer 1

此代碼根據子列表是否包含以'Derived:'開頭的字符串來分隔子列表。

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)

產量

---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? 

這段代碼最重要的部分是

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

這將迭代para （當前段落）中的各個字符串，並在找到以'Derived:'開頭的字符串時立即返回True 。

---

FWIW， for循環可以縮減為：

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

因為False和True計算為0和1，所以它們可以用來索引(therest, derived)元組。 然而，許多人會認為這是不可讀的。 :)

Answer 2

你寫的代碼幾乎是正確的。 您需要檢查列表的第3個元素中是否存在'Derived:' 。 k基本上包含paragraphs3一個元素

>>> 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'

因此，您所要做的就是將條件更改為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

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))]

詳細解釋

復制整個列表：

[l for l in paragraph3]

帶條件的復制列表：

[l for l in paragraph3 if sublist_contains('Derived: ', l)]

函數sublist_contains尚未實現，所以讓我們實現它。

僅檢索與condition_check匹配的項：

filter(condition_check, l)

由於condition_check可以表示為lambda函數：

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

將結果轉換為布爾值（如果找到至少一個匹配條件的項，則為True）：

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

並使用生成的內聯代碼替換sublist_contains ：

derived = [l for l in paragraphs3 if any(filter(lambda k: 'Derived: ' in k, l))]

Answer 4

在我看來，這似乎是最直接的方式：

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

但是，如果你願意的話，你可以獲得更多的動力，並將其拉成一條線（雖然它會比必要的更復雜）。

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

這會產生一個帶有'derived'和'therest'作為鍵的dict 。 現在你可以這樣做：

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

Answer 5

看起來你的內心清單有結構; 列表本身是一個值，而不僅僅是一個不相關的值列表。 考慮到這一點，您可以編寫一個類來表示該數據。

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中的分區配方將該列表拆分為兩個迭代器。

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)