Python 通过跳过列表之间的值将值附加到空列表

Python append values to empty list by skipping values in between a list

这只是我想要实现的目标的一个最小示例。

我有一个数组:

array = [1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,....,1,2,3,4,5,6,7,8,9,10]

我想遍历这个数组并创建一个如下所示的新数组:

new_array = [1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,....,1,2,3,4,5,6,7,8,9,10]

即遍历数组 a,获取第一个值(即 1),然后跳过 remaining 9,然后获取第一个和第二个值(即 1, 2)、然后跳过remaining 8,依此类推。

我的想法是创建 indices 并按以下方式使用它:

In [1]: indices = np.arange(1,10,1)
Out[1]: array([1, 2, 3, 4, 5, 6, 7, 8, 9])

new_array = []
for i in array:
    for a,b in zip(indices,range(10)):
        new_array.append(i[0:a]) # here I am including i[0:1], i[0:2] and so on 

所以它循环 array 并获取第一个值,然后跳过剩余的 9 个值,然后获取前两个值并跳过剩余的 8 个值,依此类推。

但这似乎不起作用。我怎样才能做到这一点?

如果你不需要所有的值(只通过你的方案)

list = [1,2,3,4,5,6,7,8,9,10] * 10
output = []

skip = 9
i = 0
while skip > 0:
    output.append(list[i])
    i += skip + 1
    skip -= 1
print(list)
print(output)

但是你的 "new_array" 没有通过你的算法。为什么不呢:

[1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10] 如果我在跳过 9 值后得到第一个 1(它有索引 0),我将得到 1,然后跳过 8 值我不会得到2

编辑: 好的,我现在明白了。这应该有效:

list = [1,2,3,4,5,6,7,8,9,10] * 10
output = []

skip = 9
i = 0
j = 0
add = 1
while skip >= 0:
    newList = list[i:j+1]
    for x in newList:
        output.append(x)
    i += skip + add
    j += skip + add + 1
    add += 1
    skip -= 1
print(output)

你能试试这个代码吗?我已经在 python 3 上测试了这段代码,它工作正常。

inp = [1,2,3,4,5,6,7,8,9,10] * 10
inp_len = len(inp);
output = [inp[0]]
skip = 9
limit= skip +2;
pointer = 1;
while skip > 0:
    pointer = pointer+skip;
    if(pointer >inp_len):
        pointer = pointer %inp_len;
    for x in inp[pointer : pointer+limit-skip ]:
        output.append(x);
    pointer= pointer+ limit-skip ;
    skip=skip-1;
print(inp)
print(output)

说明 - 添加默认的第一个元素,然后按以下顺序添加元素。

  • 跳过 9 个元素 = [1, 2]
  • 跳过 8 个元素 = [1, 2, 3]
  • 跳过 7 个元素 = [1, 2, 3, 4]
  • 跳过 6 个元素 = [1, 2, 3, 4, 5]
  • 跳过 5 个元素 = [1, 2, 3, 4, 5, 6]
  • 跳过 4 个元素 = [1, 2, 3, 4, 5, 6, 7]
  • 跳过 3 个元素 = [1, 2, 3, 4, 5, 6, 7, 8]
  • 跳过 2 个元素 = [1, 2, 3, 4, 5, 6, 7, 8, 9]
  • 跳过 1 个元素 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

请根据您的输入进行测试。我在这里使用定义的列表。

输入列表 -[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6 , 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6 , 7, 8, 9, 10]

输出列表 - [1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6 , 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10]

怎么样:

list = [1,2,3,4,5,6,7,8,9,10] * 10;
output = [];

take = 1;
index = 0;
while index < len(list):
    # take the "take" next elements, notice the "index" is not changing.
    for i in range(take):
        output.append(list[index + i]);
    # Skip the remaining values and increase the "take"
    index += 10;
    take += 1;
print(list)
print(output)

你也可以这样使用两个索引:

list = [1,2,3,4,5,6,7,8,9,10] * 10;
output = [];

index = 0;
for i in range(10):
    for j in range(10):
        if j <= i:
            output.append(list[index + i]);
        index++;
print(list)
print(output)

这与用 1 到 10 的数字打印三角形并没有太大区别。

array     = [x for x in range(1,11)]*10 #here 10 means adding same list 10 times
new_array = []
for i in range(10):
    new_array += array[10*i:10*i+i+1]
print(new_array)

希望对您有所帮助!

需要 0.0005 秒

对于单个列表,您还可以为此使用列表扩展:

list = [1,2,3,4,5,6,7,8,9,10]
output = []
i = 1
while i <= 10:
    output.extend(list[0:i])
    i +=1

print output

对于您的列表,您可以将其扩展为:

list = [1,2,3,4,5,6,7,8,9,10]*10

output = []

i = 1
j = 0
k = 1
while k <= 10:
   output.extend(list[j:i])
   j +=10
   k +=1
   i = j+k

print output   
array = [1,2,3,4,5,6,7,8,9,10]*10
new_array = []

c=j=0
while c < len(array):
    for i in range(0,j):
        new_array.append(array[i])
        i+=1
    j+=1
    c+=10
print(new_array)

输出

[1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9]