[英]convert a Python 2D list to 3D list, if there are empty elements in 2D list
Using Python3 with or without NumPy, how can I convert在有或没有 NumPy 的情况下使用 Python3,我该如何转换
a =[[1,2],[0,1],[],[2,2],[],[2,13],[1,4],[6,1],[],[2,7]]
(with respect to its empty elements), to (关于它的空元素),到
b=[[[1, 2], [0, 1]], [[2, 2]], [[2, 13], [1, 4], [6, 1]], [[2, 7]]]
I appreciate any help on this.我很感激这方面的任何帮助。
Thanks谢谢
Please try the script below:请尝试以下脚本:
a =[[],[1,2],[0,1],[],[2,2],[],[2,13],[1,4],[6,1],[],[2,7]]
newlist=[]
temp=[]
for element in a:
if element: # if element is not empty add to temp
temp.append(element)
elif temp: # if element is empty and temp is not empty add temp to newlist
newlist.append(temp)
temp=[]
if temp: # if temp is not empty add to newlist
newlist.append(temp)
print(newlist)
Using itertools's groupby
使用
itertools's groupby
from itertools import groupby
lst = []
for i, g in groupby(a, key=lambda x: len(x) >= 1):
grp = list(g)
if i and len(grp) >= 1:
lst.append(grp)
lst:第一:
[[[1, 2], [0, 1]], [[2, 2]], [[2, 13], [1, 4], [6, 1]], [[2, 7]]]
You can do this你可以这样做
def func(a):
b = [[]]
for i in (a): # iterate over a
if i: # if ith element is not empty list
b[-1].append(i) # add it to the latest list in b
else: # otherwise append new list to b
b.append([])
return b
OR Similarly,或类似地,
def func(a):
b, c = [], []
c = [b[-1].append(i) for i in ([[]] + a ) if (i or b.append([]))]
return b
Both follow basically the same process,两者都遵循基本相同的过程,
2nd snippet wastes an extra variable第二个片段浪费了一个额外的变量
You can try this.你可以试试这个。
from collections import deque
a = [[1,2],[0,1],[],[2,2],[],[2,13],[1,4],[6,1],[],[2,7]]
d = deque(a)
if d[0] == []:
d.popleft()
if len(d) > 0:
if d[-1] == []:
d.pop()
tmp = []; b = []
for x in d:
if x == []:
b.append(tmp)
tmp = []
else:
tmp.append(x)
b.append(tmp)
print(b)
outputs:输出:
[[[1, 2], [0, 1]], [[2, 2]], [[2, 13], [1, 4], [6, 1]], [[2, 7]]]
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