在python中将元素拆分为新列表

我的清单如下：

['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n', 'll Mos K 89.0\n', 'Wncy Bwn  j -97.0\n', 'May KAss S  33.58\n', 'Ai Hami s 670.0\n', 'Age Karn J 674.50\n', 'Loe AIUms s 87000.0\n', 'karl Marx J 67400.9\n', 'Joeh Stig S 34.8\n']

[' Jeoe Pmith', 'Ppseph Ksdian', ....'Joeh Stig']

['H' , 'h', 'K', .....'S']

['158.50', '590.00'....'34.8'] #(for this list, getting rid of \n as well)

它如何拆分此列表中的元素，并将其划分为3个新列表，如下所示：

['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n', 'll Mos K 89.0\n', 'Wncy Bwn  j -97.0\n', 'May KAss S  33.58\n', 'Ai Hami s 670.0\n', 'Age Karn J 674.50\n', 'Loe AIUms s 87000.0\n', 'karl Marx J 67400.9\n', 'Joeh Stig S 34.8\n']

[' Jeoe Pmith', 'Ppseph Ksdian', ....'Joeh Stig']

['H' , 'h', 'K', .....'S']

['158.50', '590.00'....'34.8'] #(for this list, getting rid of \n as well)

谢谢

L = ['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n', 'll Mos K 89.0\n', 'Wncy Bwn  j -97.0\n', 'May KAss S  33.58\n', 'Ai Hami s 670.0\n', 'Age Karn J 674.50\n', 'Loe AIUms s 87000.0\n', 'karl Marx J 67400.9\n', 'Joeh Stig S 34.8\n']

L = [s.strip().rsplit(None,2) for s in L]
first = [s[0] for s in L]
second = [s[1] for s in L]
third = [s[2] for s in L]

输出：

In [10]: first
Out[10]: 
['Jeoe Pmith',
 'Ppseph Ksdian',
 'll Mos',
 'Wncy Bwn',
 'May KAss',
 'Ai Hami',
 'Age Karn',
 'Loe AIUms',
 'karl Marx',
 'Joeh Stig']

In [11]: second
Out[11]: ['H', 'h', 'K', 'j', 'S', 's', 'J', 's', 'J', 'S']

In [12]: third
Out[12]: 
['158.50',
 '590.00',
 '89.0',
 '-97.0',
 '33.58',
 '670.0',
 '674.50',
 '87000.0',
 '67400.9',
 '34.8']

输出：

In [10]: first
Out[10]: 
['Jeoe Pmith',
 'Ppseph Ksdian',
 'll Mos',
 'Wncy Bwn',
 'May KAss',
 'Ai Hami',
 'Age Karn',
 'Loe AIUms',
 'karl Marx',
 'Joeh Stig']

In [11]: second
Out[11]: ['H', 'h', 'K', 'j', 'S', 's', 'J', 's', 'J', 'S']

In [12]: third
Out[12]: 
['158.50',
 '590.00',
 '89.0',
 '-97.0',
 '33.58',
 '670.0',
 '674.50',
 '87000.0',
 '67400.9',
 '34.8']

那么：

>>> l = ['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n', 'll Mos K 89.0\n', 'Wncy Bwn  j -97.0\n', 'May KAss S  33.58\n', 'Ai Hami s 670.0\n', 'Age Karn J 674.50\n', 'Loe AIUms s 87000.0\n', 'karl Marx J 67400.9\n', 'Joeh Stig S 34.8\n']
>>> zip(*(el.rsplit(None, 2) for el in l))
[('Jeoe Pmith', 'Ppseph Ksdian', 'll Mos', 'Wncy Bwn', 'May KAss', 'Ai Hami', 'Age Karn', 'Loe AIUms', 'karl Marx', 'Joeh Stig'), ('H', 'h', 'K', 'j', 'S', 's', 'J', 's', 'J', 'S'), ('158.50', '590.00', '89.0', '-97.0', '33.58', '670.0', '674.50', '87000.0', '67400.9', '34.8')]

它给出的是一个元组列表，而不是列表列表，但如果您关心它，这很容易更改。

如何：

>>> l = ['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n', 'll Mos K 89.0\n', 'Wncy Bwn  j -97.0\n', 'May KAss S  33.58\n', 'Ai Hami s 670.0\n', 'Age Karn J 674.50\n', 'Loe AIUms s 87000.0\n', 'karl Marx J 67400.9\n', 'Joeh Stig S 34.8\n']
>>> zip(*(el.rsplit(None, 2) for el in l))
[('Jeoe Pmith', 'Ppseph Ksdian', 'll Mos', 'Wncy Bwn', 'May KAss', 'Ai Hami', 'Age Karn', 'Loe AIUms', 'karl Marx', 'Joeh Stig'), ('H', 'h', 'K', 'j', 'S', 's', 'J', 's', 'J', 'S'), ('158.50', '590.00', '89.0', '-97.0', '33.58', '670.0', '674.50', '87000.0', '67400.9', '34.8')]

---

它提供了一个元组列表，而不是列表列表，但如果您关心它，这很容易更改。

最基本的解决方案，仅供参考：

lines = ['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n']

list1 = []
list2 = []
list3 = []

for line in lines:
    cleaned = line.strip()  # drop the newline character
    splitted = cleaned.rsplit(' ', 2)  # split on space 2 times from the right
    list1.append(splitted[0])
    list2.append(splitted[1])
    list3.append(splitted[2])

print list1, list2, list3

---

最基本的解决方案，仅供参考：

lines = ['Jeoe Pmith H 158.50\n', 'Ppseph Ksdian h 590.00\n']

list1 = []
list2 = []
list3 = []

for line in lines:
    cleaned = line.strip()  # drop the newline character
    splitted = cleaned.rsplit(' ', 2)  # split on space 2 times from the right
    list1.append(splitted[0])
    list2.append(splitted[1])
    list3.append(splitted[2])

print list1, list2, list3

---

非常感谢。但是你能解释一下这行代码吗？L=[s.strip.rsplitNone，2代表s在L]谢谢！！！！但是你能解释一下这行代码吗？L=[s.strip.rsplitNone，2代表L中的s]为什么说这是最基本的解决方案？它涉及到与其他人一样的分裂。不要使用列表理解，而是有一个foor循环。您还必须声明三个列表。@RaydelMiranda正是因为它没有列表理解或zip和splice。这使它成为一个基本的解决方案。你为什么说这是最基本的解决方案？它涉及到与其他人一样的分裂。不要使用列表理解，而是有一个foor循环。您还必须声明三个列表。@RaydelMiranda正是因为它没有列表理解或zip和splice。这使它成为一个基本的解决方案。