将具有连续块项目但不连续块的列表拆分为子列表

Question

假设我有一个这样的列表：

mylist = [2,3,4,7,8,9,0,1,2,3,10,11,12,20,19,18,17,17,18,19,20,21]

我想将其分成这样的子列表：

sublist_1 = [2,3,4]
sublist_2 = [7,8,9]
sublist_3 = [0,1,2,3]
sublist_4 = [10,11,12]
sublist_5 = [20,19,18,17]
sublist_5 = [17,18,19,20,21]

我怎样才能做到这一点？

Answer 1

代码-

mylist = [2, 3, 4, 7, 8, 9, 0, 1, 2, 3, 10, 11, 12, 20, 19, 18, 17, 17, 18, 19,
          20, 21]

result = []

temp_list = [mylist[0]]

for i in range(1, len(mylist)):
    if abs(mylist[i] - mylist[i - 1]) != 1:
        result.append(temp_list)
        temp_list = [mylist[i]]
    else:
        temp_list.append(mylist[i])

if temp_list:
    result.append(temp_list)

print(result)

输出-

[[2, 3, 4], [7, 8, 9], [0, 1, 2, 3], [10, 11, 12], [20, 19, 18, 17],
 [17, 18, 19, 20, 21]]

Answer 2

这样就可以了；

mylist = [2,3,4,7,8,9,0,1,2,3,10,11,12,20,19,18,17,17,18,19,20,21]

index =0
for i in range(0, len(mylist)):
    #print (mylist[i+1])
    if (i+1 >= len(mylist)):
        print (mylist[index:i+1])
        break

    if not( (mylist[i+1] - mylist[i] == 1) or (mylist[i+1] - mylist[i] == -1) ):
        print (mylist[index:i+1])
        index = i+1

产量

[2, 3, 4]
[7, 8, 9]
[0, 1, 2, 3]
[10, 11, 12]
[20, 19, 18, 17]
[17, 18, 19, 20, 21]

Answer 3

这是numpy方法。 我正在使用ediff1d ， intersect1d ， 非零

import numpy as np
mylist = [2, 3, 4, 7, 8, 9, 0, 1, 2, 3, 10, 11, 12, 20, 19, 18, 17, 17, 18, 19,
          20, 21]
a = np.array(mylist)
t = np.ediff1d(mylist)
i = np.intersect1d(np.nonzero(t!=-1), np.nonzero(t!=1))
rslt = np.split(a,i+1)


rslt
Out[70]: 
[array([2, 3, 4]),
 array([7, 8, 9]),
 array([0, 1, 2, 3]),
 array([10, 11, 12]),
 array([20, 19, 18, 17]),
 array([17, 18, 19, 20, 21])]

如果要输出为list而不是numpy array ，请将最后一行更改为rslt = [x.tolist() for x in np.split(a,i+1)] ，结果将是

rslt
Out[9]: 
[[2, 3, 4],
 [7, 8, 9],
 [0, 1, 2, 3],
 [10, 11, 12],
 [20, 19, 18, 17],
 [17, 18, 19, 20, 21]]