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)
(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:
('\n'.join(i))
('\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(('Derived:') for item in para):
(para)
else:
(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(('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(('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. ????