Python列表理解具有可变长度的多个元素？

I'm writing code to convert a CSV into XML. Suppose I have a single list like:

我正在编写代码将CSV转换为XML。假设我有一个列表，如：

input = ['name', 'val', 0, \
         'name', 'val', 1, 'tag', 'val', \
         'name', 'val', 2, 'tag', 'val', 'tag', 'val', \
         'name', 'val', 0]

Every slice of this list starting with 'name' denotes an element with a name, a value, and a variable number of optional tag-value pairs.

此列表的每个切片以“name”开头，表示具有名称，值和可变标记值对的可变数量的元素。

I want to turn this into:

我想把它变成：

output = [['name', 'val', []], 
          ['name', 'val', ['tag', 'val']],
          ['name', 'val', ['tag', 'val', 'tag', 'val']],
          ['name', 'val', []]]

No need to separate the tag-value pairs into tuples, this is handled in a separate method. I have a solution, but it's not very pythonic:

无需将标记值对分隔为元组，这是在单独的方法中处理的。我有一个解决方案，但它不是非常pythonic：

output=[]
cursor=0

while cursor < len(input):
    name=input[cursor]
    val=input[cursor+1]
    ntags=int(input[cursor+2])
    optslice=input[cursor+3:cursor+3+2*ntags]
    cursor = cursor+3+2*ntags
    print name, val, ntags, optslice, cursor
    output.append([name, val, optslice])    
print output

> name val 0 [] 3
> name val 1 ['tag', 'val'] 8
> name val 2 ['tag', 'val', 'tag', 'val'] 15
> name val 0 [] 18

> [['name', 'val', []], ['name', 'val', ['tag', 'val']], ['name', 'val', ['tag', 'val', 'tag', 'val']], ['name', 'val', []]]

I think I could probably do this as a list comprehension, but the variable length of each element is throwing me for a loop. The input is parsed from a CSV, and I can change the format to better fit a different solution. Ideas?

我想我可能会把它作为一个列表理解，但是每个元素的可变长度都会让我陷入循环。输入是从CSV解析的，我可以更改格式以更好地适应不同的解决方案。想法？

4 个解决方案

#1

I'd use an iterator instead of your cursor and then drive the comprehension with for name in it.

我将使用迭代器而不是光标，然后使用名称驱动理解。

it = iter(input)
output = [[name, next(it), [next(it) for _ in range(2 * next(it))]] for name in it]

Or with islice:

或者使用islice：

from itertools import islice

it = iter(input)
output = [[name, next(it), list(islice(it, 2 * next(it)))] for name in it]

That said, I suspect you shouldn't have all data in that flat list in the first place. Likely your CSV file has structure that you should use instead. I.e., don't flatten two-dimensional data just so you need to unflatten it back. But your question is interesting nonetheless :-)

也就是说，我怀疑你不应该首先在那个平面列表中包含所有数据。可能您的CSV文件具有您应该使用的结构。即，不要将二维数据展平，因此您需要将其展开。但你的问题很有意思:-)

#2

I don't know how pythonic you consider this, but you could do something like this

我不知道你认为pythonic是多么神圣，但你可以这样做

finallist = []
therest = x

while therest:
    name, val, count, *therest = therest
    sublist, therest = rest[:2*count], rest[2*count:]
    finallist.append([name, val] + [sublist])

#3

Here is my code:

这是我的代码：

data = ['name', 'val', 0,
        'name', 'val', 1, 'tag', 'val',
        'name', 'val', 2, 'tag', 'val', 'tag', 'val',
        'name', 'val', 0]

tmp = [
    [
        data[pos:pos + 2],
        [i for i in data[pos + 3:pos + 3 + data[pos + 2] * 2]]
    ] for pos, e in enumerate(data) if e == 'name']

for e in tmp:
    print e

The output is:

输出是：

# [['name', 'val'], []]
# [['name', 'val'], ['tag', 'val']]
# [['name', 'val'], ['tag', 'val', 'tag', 'val']]
# [['name', 'val'], []]

#4

If you really want to use pure list comprehension:

如果你真的想使用纯列表理解：

a = ['name', 'val', 0, \
              'name', 'val', 1, 'tag', 'val', \
              'name', 'val', 2, 'tag', 'val', 'tag', 'val', \
              'name', 'val', 0]


print(
[grouped[:2] + [tag for tag in grouped[3:]] for grouped in
    [
        a[i:i+(a[i+1:].index("name") + 1 if a[i+1:].count("name") else len(a[i:])+1)]
        for i, x in enumerate(a) if x == "name"
    ]
])

It is really ugly though.

这真的很难看。

#1