Question

我有这段代码：

from datetime import datetime

values = [[datetime(2017, 1, 22, 10, 2, 11, 170913), datetime(2017, 1, 22, 10, 3, 11, 170913), 
           datetime(2017, 1, 2, 10, 4, 11, 170913), datetime(2017, 1, 22, 10, 5, 11, 170913)],
          [datetime(2017, 1, 22, 10, 1, 11, 170913), datetime(2017, 1, 22, 10, 2, 11, 170913), 
           datetime(2017, 1, 22, 10, 3, 11, 170913), datetime(2017, 1, 22, 10, 4, 11, 170913)]]

secs = []
secs.append([i.timestamp() for t in values for i in t])
print(secs)

输出结果为：

[[1485072131.170913, 1485072191.170913, 1483344251.170913, 1485072311.170913, 1485072071.170913, 1485072131.170913, 1485072191.170913, 1485072251.170913]]

我希望输出类似于输入（值）形状：

[[1485072131.170913, 1485072191.170913, 1483344251.170913, 1485072311.170913], 
[1485072071.170913, 1485072131.170913, 1485072191.170913, 1485072251.170913]]

我知道我可以对结果进行重塑，但是当我追加时，还有另一种方法吗？

Answer 1

简单地使列表理解嵌套而不是平坦：

secs = [[i.timestamp() for i in t] for t in values]
print(secs)

你会得到你想要的东西：

[[1485097331.170913, 1485097391.170913, 1483369451.170913, 1485097511.170913], 
[1485097271.170913, 1485097331.170913, 1485097391.170913, 1485097451.170913]]

将datetime对象追加到列表中，按形状分组

1 个答案: