python – 在几乎相等的部分和返回边界中拆分列表

发布时间：2020-12-20 11:04:07 所属栏目：Python 来源：网络整理

导读：我想拆分列表并返回边界,因此在扩展中不应该存在单个元素. 例如 split(list(range(1,101)),2) # should return [[1,50],[51,100]]split(list(range(1,3)# should return[[1,33],[34,66],[67,6)),2],[3,5]] # Ideally last element should merge with last if

我想拆分列表并返回边界,因此在扩展中不应该存在单个元素.

例如

split(list(range(1,101)),2) 
# should return 
[[1,50],[51,100]]

split(list(range(1,3)
# should return
[[1,33],[34,66],[67,6)),2],[3,5]] # Ideally last element should merge with last if last one has no pair.

到目前为止我试过了

def split(l,n):
    x = list(range(1,l+1))
    return [x[i:i+n] for i in range(0,len(x),int(len(x)/n))]

print(split(20,2))

返回[[1,[11,12]]而不是[[1,10],20]]

解决方法

def csplit(m,n):
    div_ = m//n
    step = div_ if div_ > 1 else 2  # determine step for range function (at least 2 'alternate steps')
    lis = []
    for i in range(1,m+1,step):
        if (m-(i+max(1,div_-1))) > 1:
            # append list only if remains at least two elements remains 
            lis.append([i,i+max(1,div_-1)])
        else:
            if not m == i:
                # in case if m and i not equal and not more then one element left then construct list which include that element
                lis.append([i,m])
            break  # break the loop from iterating any further
    return lis

if __name__ == "__main__":
    print(csplit(100,2))
    print(csplit(100,3))
    print(csplit(5,3))

输出：

[[1,100]]
[[1,5]]

同一个衬垫：

def csplit(m,n):
    return [[i,m//n-1)] if (m-(i+max(1,m//n-1))) > 1 else [i,m] for i in range(1,max(m//n,2)) if not i==m]

（编辑：李大同）

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