prev_day
=
datetime
.
datetime
(
2017
,
12
,
31
)
cur_day
=
datetime
.
datetime
(
2018
,
1
,
31
)
diff
=
cur_day
-
prev_day
print
(
diff
)
31 days, 0:00:00
diff是datetime.timedelta类型的数据
datetime.timedelta类型转换为int类型：diff.days
import datetime
prev_day = datetime.datetime(2017, 12, 31)
cur_day = datetime.datetime(2018, 1, 31)
diff = cur_day - prev_day
print(diff)
print(diff.days)
print(type(diff.days))
31 days, 0:00:00
<class 'int'>
import datetimeprev_day = datetime.datetime(2017, 12, 31)cur_day = datetime.datetime(2018, 1, 31)diff = cur_day - prev_dayprint(diff)31 days, 0:00:00diff是datetime.timedelta类型的数据datetime.time...
#选取并构建LRFMC模型的特征
airline_selection = airline[["FFP_DATE","LOAD_TIME","FLIGHT_COUNT","LAST_TO_END",
"avg_discount","SEG_KM_SUM"]] #筛选多个列要用两个中括号
#构建L特征
L = pd....
要将一个时间序列中的datetime类型数据从"2 days"形式转换为int类型，可以使用pandas的dt属性和total_seconds方法。首先，将时间序列转换为timedelta类型，然后使用dt.total_seconds方法计算总秒数，最后将结果转换为int类型。
示例代码如下：
import pandasas pd
# 创建时间序列
times = pd.Series(["2 d...
15 NaT
Name: onset_days, dtype: timedelta64[ns]
Name: onset_days, dtype: timedelta64[ns]
格式为timedelta64[ns]
首先将缺失
today = datetime.datetime.today()
print('当前时间：',today)
print('当前时间转换成整h整m整s:',today.replace(minute=0, second=0))
# 时间的加减
res1 = today + datetime.timedelta(days=1,minutes=60)
print('增加的时间为res:',res1)
res0 = datetime.datetime.now()
print('未格式化时间res0：',res0)
long t;
t = DateTime.Now.Ticks; textBox1.Text = t.ToString();
textBox2.Text = DateTime.FromFileTimeUtc(t).ToString();
转载于:https://www.cnblogs.com/zyh-C/p/10084433...
datetime 模块——timedelta类;timedelta类：datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, hours=0, weeks=0)
timedelta类是用来计算二个datetime对象的差值的。此类中包含如下属性：
1、days:天数
2、microseconds：微秒数(>=0 并且 <1秒）
3、seconds：秒数(>=0 并且 <1天）;timedelta类;timedelta类
