Question

我有一个列表列表：

ex = [['1001'],['0010'],['1101'],['0000']]

我想将此列表列表拆分为较小的列表。我还有另一个列表，其中包含要分割的索引：

track = [1,3]

所以我想分割这个列表列表，以得到以下结果：

sublist = [
    [[1,0],[0,0],[1,1],[0,0]],
    [[0,1],[1,0],[0,1],[0,0]]
    ]

我已经尝试了一个简单的列表：

ex = [1,0,0,1]
start = 0
position = []
for i in track:
    position.append(ex[start:i+1])
    start = i+1

但是在这种情况下，我的列表已经有整数，而原始列表有字符串。

如何在具有字符串而不是整数的列表列表上实现此目的？我不知道从哪里开始？

Answer 1

您似乎想将数字拆分为数字，然后将每个数字的前两位和后两位放入结果的单独列表中-您可以进行以下操作：

# what you got
ex = [['1001'],['0010'],['1101'],['0000']]

# what you want
sublist = [[[1,0],[0,0],[1,1],[0,0]],
           [[0,1],[1,0],[0,1],[0,0]]]

# how to get there: create single integers from each string
# list comprehension, see below for answers about them
digits = [ list(map(int,l)) for inner in ex for l in inner] 
print(digits )

# create the results 
result = [ [],[] ]

for inner in digits:
    result[ 0].append( inner[:2] )   # list slicing, see below for answers about it
    result[-1].append( inner[2:] )

print(result)

输出（重新格式化）：

# split into digits
[[1, 0, 0, 1], [0, 0, 1, 0], [1, 1, 0, 1], [0, 0, 0, 0]]

# put into results
[[[1, 0], [0, 0], [1, 1], [0, 0]],
 [[0, 1], [1, 0], [0, 1], [0, 0]]]

Built in functions可帮助您解释map()和其他有用的功能。读起来也很有趣：

Answer 2

sublist是最终列表。 subsublist被视为中间临时列表。
track是外部循环
ex是内循环

ex = [['1001'],['0010'],['1101'],['0000']]
track = [1,3]
subsublist = []
sublist = []
start=0

for index in track:
    # print(index)
    if start == 0:
        end=index+1
    else:
        end=None
    for item in ex:
        subsublist.append([item[0][start:end]])
    sublist.append(subsublist)
    start=end
    subsublist = []

print(sublist)

[UPDATE]另一种使代码通用的尝试！

ex = [['100190'],['001099'],['110187'],['000050']]
tracks = [1,3,6]
subsublist = []
sublist = []
start=0

for track in tracks:
    indexOfTrack = tracks.index(track)
    if indexOfTrack == 0:
        end=track+1
    elif indexOfTrack > 0 and indexOfTrack < len(tracks)-1:
        end = track+1
    else:
        end=None
    for item in ex:
        subsublist.append([item[0][start:end]])
    sublist.append(subsublist)
    start=end
    subsublist = []

print(sublist)

Answer 3

import itertools
import sys

ex = [['1001'], ['0010'], ['1101'], ['0000']]
track = [1, 3]

# [a,b)
it = itertools.chain(map(lambda x: x - 1, track), [sys.maxsize])
last = next(it, None)
result = []
for curr in it:
    temp = []
    for s in itertools.chain.from_iterable(ex):
        temp.append(list(map(int, s[last:curr])))
    result.append(temp)
    last = curr

print(result)

将列表列表拆分为较小的列表

3 个答案: