按某些重复索引值拆分列表

Question

我有一个整数列表，其中一些是连续的数字。

是）我有的：

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]等...

我想要的是：

MyNewIntList = [[21,22,23,24],[0,1,2,3],[0,1,2,3,4,5,6,7]]

我希望能够通过元素0拆分此列表，即在循环时，如果元素为0，则将列表拆分为单独的列表。 然后，在分割myIntList任意次数（基于找到元素0的重复）之后，我想将每个“split”或一组连续整数附加到列表中的列表中。

我还可以使用'字符串列表'而不是整数来做同样的事情吗？ （根据重复出现的元素将主字符串列表拆分为较小的列表）

编辑：

我如何按连续数字拆分列表？ 我的列表中有一部分从322跳到51，中间没有0。 我想拆分：

[[...319,320,321,322,51,52,53...]]

成

[[...319,320,321,322],[51,52,53...]]

基本上，如何按连续数字拆分列表中的元素？

发布在此处： 按顺序拆分列表（整数）到单独的列表中

Answer 1

it  = iter(myIntList)
out = [[next(it)]]
for ele in it:
    if ele != 0:
        out[-1].append(ele)
    else:
        out.append([ele])

print(out)

或者在一个功能中：

def split_at(i, l):
    it = iter(l)
    out = [next(it)]
    for ele in it:
        if ele != i:
            out.append(ele)
        else:
            yield out
            out = [ele]
    yield out

如果你在开始时有0 ，它会被捕获：

In [89]: list(split_at(0, myIntList))
Out[89]: [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

In [90]: myIntList = [0,21, 22, 23, 24, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7]

In [91]: list(split_at(0, myIntList))
Out[91]: [[0, 21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

Answer 2

（我隐约怀疑我以前做过这个，但我现在找不到它。）

from itertools import groupby, accumulate

def itergroup(seq, val):
    it = iter(seq)    
    grouped = groupby(accumulate(x==val for x in seq))
    return [[next(it) for c in g] for k,g in grouped]

给

>>> itergroup([21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7], 0)
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]
>>> itergroup([0,1,2,0,3,4], 0)
[[0, 1, 2], [0, 3, 4]]
>>> itergroup([0,0], 0)
[[0], [0]]

（也就是说，在实践中我使用与其他人相同的循环/分支的yield版本，但是我会发布上面的变种。）

Answer 3

你可以使用切片：

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
myNewIntList = []
lastIndex = 0
for i in range(len(myIntList)):
    if myIntList[i] == 0:
        myNewIntList.append(myIntList[lastIndex:i])
        lastIndex = i

myNewIntList.append(myIntList[lastIndex:])
print(myNewIntList)
# [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

您可以使用str.split函数拆分字符串：

s = 'stackoverflow'
s.split('o') # ['stack', 'verfl', 'w'] (removes the 'o's)

import re
[part for part in re.split('(o[^o]*)', s) if part] # ['stack', 'overfl', 'ow'] (keeps the 'o's)

Answer 4

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]

new = []
m,j=0,0
for i in range(myIntList.count(0)+1):
    try:
        j= j+myIntList[j:].index(0)
        if m==j:
           j= j+myIntList[j+1:].index(0)+1



        new.append(myIntList[m:j])
        m,j=j,m+j
    except:
        new.append(myIntList[m:])
        break
print new

产量

 [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

OUTPUT2

myIntList = [0,21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]

[[0, 21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

Answer 5

您可以遍历整个列表，附加到临时列表，直到找到0 。 然后再次重置临时列表并继续。

>>> myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
>>> newlist = [] 
>>> templist = []
>>> for i in myIntList:
...      if i==0:
...          newlist.append(templist)
...          templist = []
...      templist.append(i)
... 
>>> newlist.append(templist)
>>> newlist
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

对于字符串，您可以使用list调用使用相同的方法

>>> s = "winterbash"
>>> list(s)
['w', 'i', 'n', 't', 'e', 'r', 'b', 'a', 's', 'h']

也使用itertools

>>> import itertools
>>> myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
>>> temp=[list(g) for k,g in itertools.groupby(myIntList,lambda x:x== 0) if not k]
>>> if myIntList[0]!=0:
...     newlist = [temp[0]] + [[0]+i for i in temp[1:]]
... else:
...     newlist = [[0]+i for i in temp]
... 
>>> newlist
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

Answer 6

你可以尝试：

i = 0
j = 0
loop = True
newList = []

while loop:
    try:
        i = myIntList.index(0, j)
        newList.append(myIntList[j:i])
        j = i + 1
    except ValueError as e:
        newList.append(myIntList[j:])
        loop = False

print newList
[[21, 22, 23, 24], [1, 2, 3], [1, 2, 3, 4, 5, 6, 7]]