How to split one list in a list of x list with python?
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如何在python中拆分x列表列表中的列表？
例如，列表分为两个列表：
1 2 | elements = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] -> ([0, 1, 2, 3, 4], [5, 6, 7, 8, 9]) |
示例：列表分为3个列表(注意第三个元组列表，包含"division rest")：
1 2 | elements = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] -> ([0, 1, 2], [3, 4, 5], [6, 7, 8, 9]) |
等。。。
您可以这样尝试：
编辑2:
1 2 3 4 5 6 7 8 9 | i,j,x=len(seq),0,[] for k in range(m): a, j = j, j + (i+k)//m x.append(seq[a:j]) return x |
像呼叫
1 | seq = range(11), m=3 |
结果
1 | result : [[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]] |
1 2 | def chunks(l, n): return [l[i:i + n] for i in range(0, len(l), n)] |
Eddi1:
1 2 3 | >>> x = [1,2,3,4,5,6,7,8,9] >>> zip(*[iter(x)]*3) [(1, 2, 3), (4, 5, 6), (7, 8, 9)] |
1 2 3 4 5 6 7 8 9 10 | def chunk(iterable ,n): le = len(iterable) if n > le: return iterable mo = le % n # get mod from length of the iterable % required len sublists diff = le - mo # if there was any mod, i.e 10 % 3 = 1, diff will be 9 sli = le / n res = [iterable[ind:ind + sli] for ind in range(0, diff, sli)] res[-1] = res[-1] + iterable[diff:] # add from diff to end of the last sublist return tuple(res) # return a tuple lists |
如果