重复列表的序列

Question

问题

给定以下列表：

l=[1,2,3,4,5,6]

我想重复 n 列表的第一个元素 r 次，然后，对列表的下一个 n 元素执行相同的操作，依此类推...

例如，对于 n=2 和 r=2：

result=[1,2,1,2,3,4,3,4,5,6,5,6]

对于n=3和r=2：

result=[1,2,3,1,2,3,4,5,6,4,5,6]

我的拙劣尝试

l=[1,2,3,4,5,6]
n=3
r=2
l_new=[]
result=[]
for i in range(0,len(l),n):
    reduced=l[i:(n+i)]
    l_new.append(reduced)

for elem in l_new:
    elem_new=elem*r
    result+=(elem_new)

我该如何改进这段代码？ clean/efficient/good 更实用的做法是什么？

Answer 1

你可以像下面这样干净利落地实现。

l=[1,2,3,4,5,6]

def repeat_n_r_times(l, n, r):
    list_breakdown = [l[i:i+n] for i in range(0, len(l), n)]
    #make copies r times 
    lst = [lst*r for lst in list_breakdown]
    # merge all sublist
    return [item for sublist in lst for item in sublist]

print(repeat_n_r_times(l, 3, 2))
# OUTPUT: [1,2,3,1,2,3,4,5,6,4,5,6]
print(repeat_n_r_times(l, 2, 2))
# OUTPUT: result=[1,2,1,2,3,4,3,4,5,6,5,6]

Answer 2

像这样的东西应该适合你：

l=[1,2,3,4,5,6]
n=3
r=2
result=[]

for i in range(0,len(l),n):
    result += l[i:i+n]*r

或者作为一个函数：

def repeat_sequence(l, n, r):
    result = []

    for i in range(0,len(l),n):
        result += l[i:i+n]*r

    return result

l=[1,2,3,4,5,6]
repeat_sequence(l, 3, 2)

Answer 3

def flatten(arr):
    return [item for sublist in arr for item in sublist]

def flatmap(f, arr):
    return flatten(map(f, arr))

def chunk(arr, r):
    return [arr[i:i+r] for i in range(0,len(arr), r)]

def repeat_n_r_times(arr, n, r):
  return flatmap(lambda x: x * n, chunk(arr, r))

重复列表的序列

Repeat a sequence of a list

python

loops

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combinatorics