[已解决]将多位整数转成列表

jtxs0000

# 第一个问题
# nums = [2,5,3,8]

# 输出为：[5,5,8,8,8]  2个5，3个8


# # 第二个问题
# nums = [42,39]        
# 输出为：[39,39,39,39,39,39,39,39,39,39,39,39,39......]   一共42个39

第一个基本会做了，但用第一个的代码运行第二个会先显示

['3', '9', '3', '9', '3', '9', '3', '9', '3', '9', '3', '9', '3', '9']

有没有大佬会这个的

最佳答案

月排行榜 / 总排行榜

疾风怪盗

2020-10-13 15:15:25

jtxs0000 发表于 2020-10-13 15:07
好吧，应该是我没有描述清楚，我是想要写一组代码，能把第一个问题和第二个问题都能运行起来，甚至说，如 ...

这样？

nums = [42,39,12,5,100,23]
new=[nums[i+1] for i in range(0,len(nums),2) for a in range(1,nums[i]+1)]
print(new)

[39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23]

跳转到最佳答案楼层

冬雪雪冬

>>> nums = [42,39]
>>> print([nums[1]] * nums[0])
[39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39]

jtxs0000

冬雪雪冬 发表于 2020-10-13 15:03

好吧，应该是我没有描述清楚，我是想要写一组代码，能把第一个问题和第二个问题都能运行起来，甚至说，如果出现其他类似的也可以运行，比如

nums = [42,39，12，5，100，23]

疾风怪盗

jtxs0000 发表于 2020-10-13 15:07
好吧，应该是我没有描述清楚，我是想要写一组代码，能把第一个问题和第二个问题都能运行起来，甚至说，如 ...

这样？

nums = [42,39,12,5,100,23]
new=[nums[i+1] for i in range(0,len(nums),2) for a in range(1,nums[i]+1)]
print(new)

[39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 39, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23]

冬雪雪冬

jtxs0000 发表于 2020-10-13 15:07
好吧，应该是我没有描述清楚，我是想要写一组代码，能把第一个问题和第二个问题都能运行起来，甚至说，如 ...

nums = [42, 39, 12, 5, 100, 23]
output = []
for each in range(0, len(nums), 2):
    output += nums[each] * [nums[each + 1]]
print(output)

伪文青

num = [5,3,2,4,2,8]
l = len(num)
nums = []
if l%2==0:
    #偶数输出列表
    for i in range(0,l):
        if i%2!=0:
            nums.extend([num[i]] * num[i-1])
else:
    #奇数报错
    print("少一个数字")
print(nums)