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[英]Python . How to sort elements of sub-list if second elements are equal sort the sub-lists by first element
[英]How to create sub-lists with 3 elements, but if a sub-list gets a single element then instead create two with 2 elements?
以下示例阐明了 2 元素子列表用例:
如果一个列表有 10 个元素[A, B, C, D, E, F, G, H, I, J]
我想得到 4 个子列表: [[A, B, C], [D, E, F], [G, H], [I, J]]
。
提前致谢。
怎么样:
l = list('ABCDEFGHIJ')
n = len(l)
res = [ l[(i if i < n-1 else i-1) : (i+3 if i < n-4 or i == n-3 else i+2)] for i in range(0,n,3)]
print(res)
这将给出以下输出:
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I']]
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H'], ['I', 'J']]
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I'], ['J', 'K']]
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I'], ['J', 'K', 'L']]
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I'], ['J', 'K'], ['L', 'M']]
这是使用迭代器的解决方案:
L = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
N = len(L)
I = iter(L)
[[next(I) for _ in range(size)] for size in [3]*(N//3-1) + ([3] if (N%3)==0 else ([2,2] if (N%3)==1 else [3,2]))]
[['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H'], ['I', 'J']]
我会做一个 function:
def partition(a):
arr = np.array(a)
n = len(arr) # assume that n>=3
r = n % 3
if r == 0:
return list(arr.reshape(-1, 3))
if r == 1:
return list(arr[:-4].reshape(-1,3)) + list(arr[-4:].reshape(-1,2))
# r == 2
return list(arr[:-2].reshape(-1,3)) + [arr[-2:]]
测试:
partition(list('ABCDEFGHIJ'))
# [array(['A', 'B', 'C'], dtype='<U1'), array(['D', 'E', 'F'], dtype='<U1'),
# array(['G', 'H'], dtype='<U1'), array(['I', 'J'], dtype='<U1')]
partition(list('ABCDEFGHI'))
# [array(['A', 'B', 'C'], dtype='<U1'), array(['D', 'E', 'F'], dtype='<U1'),
# array(['G', 'H', 'I'], dtype='<U1')]
partition(list('ABCDEFGH'))
# [array(['A', 'B', 'C'], dtype='<U1'), array(['D', 'E', 'F'], dtype='<U1'),
# array(['G', 'H'], dtype='<U1')]
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