将我从下面的查询中获得的 90'000 个 datetime.datetime 对象的元组转换为自午夜以来的秒数元组的最佳方法是什么？
例如2014-06-13 10:33:20 应变为 38000 (10*3600+33*60+20)
query = """
SELECT timestamp, value
FROM measurements
WHERE timestamp BETWEEN %s AND %s AND sensor = %s
"""
cursor.execute(query, (start, stop, sensor))
row = cursor.fetchall()
timestamp, value=zip(*row)
最佳答案
创建一个函数来转换为秒，然后使用 list comprehension .
from datetime import datetime, timedelta
import random
# Generate some random datetime objects.
d = [datetime.today() + timedelta(seconds=i*600) for i in range(10)]
def dt_to_seconds(dt):
return 3600*dt.hour + 60*dt.minute + dt.second
s = tuple([dt_to_seconds(i) for i in d])
print(s)
# (35545, 36145, 36745, 37345, 37945, 38545, 39145, 39745, 40345, 40945)
关于python - 日期时间元组到秒元组，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24201087/