在python中包含字符串的列表中提取列表

2023年6月3日 266次阅读

我正在尝试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 = []最终在where = [],除非我做这样的事情：

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？我正在寻找的最终结果/输出是：

Page: 2
Bib: Something
Derived: This n that

Page: 3
Bib: Something
.
.
.

最佳答案 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)

产量

---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?

这段代码最重要的部分是

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

这将迭代para(当前段落)中的各个字符串,并在找到以“Derived：”开头的字符串时立即返回True.

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,所以它们可以用来索引(其中,派生的)元组.然而,许多人会认为这是不可读的. 🙂

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