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2018-08-05 17:45:20 +08:00 回复了 Laccuse 创建的主题 › Python › 请问 Python 如何实现两个数字一行，三个数字一行交替输出? |
获取到 4 位数中能被 3 整除不能被 6 整除的数
list = filter(None, [i if(i%3 is 0 and i%6 is not 0) else None for i in range(1000, 10000)])
写一个返回指定长度并从元列表中剔除这些元素的方法
def push_num(list, reqlen):
count = 0
req = []
while count is not reqlen:
req.append(list[0])
del list[0]
count += 1
return req
再写一个交替判断
def list_slice(list):
req = []
req.append(push_num(list, 2))
while True:
if len(list) is 0:
return req
if len(req[-1]) is 3:
req.append(push_num(list, 2))
elif len(req[-1]) is 2:
req.append(push_num(list, 3))
最后 结合一下
list_slice( filter(None, [i if(i%3 is 0 and i%6 is not 0) else None for i in range(1000, 10000)]) )
list = filter(None, [i if(i%3 is 0 and i%6 is not 0) else None for i in range(1000, 10000)])
写一个返回指定长度并从元列表中剔除这些元素的方法
def push_num(list, reqlen):
count = 0
req = []
while count is not reqlen:
req.append(list[0])
del list[0]
count += 1
return req
再写一个交替判断
def list_slice(list):
req = []
req.append(push_num(list, 2))
while True:
if len(list) is 0:
return req
if len(req[-1]) is 3:
req.append(push_num(list, 2))
elif len(req[-1]) is 2:
req.append(push_num(list, 3))
最后 结合一下
list_slice( filter(None, [i if(i%3 is 0 and i%6 is not 0) else None for i in range(1000, 10000)]) )
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