Question

我有df['Status']列，其中包含一些对象：

In: df.Status.unique() Out: array([nan, 'Open', 'Plmt', 'SHRT', 'Check'], dtype=object)

列：

In: df['Status'] Out: time Status 2016-01-15 08:55:00 Open 2016-01-15 09:00:00 Plmt 2016-01-15 09:05:00 Plmt 2016-01-15 09:10:00 Plmt 2016-01-15 09:15:00 Plmt 2016-01-15 09:20:00 Plmt 2016-01-15 09:25:00 Plmt 2016-01-15 09:30:00 Plmt 2016-01-15 09:35:00 Plmt 2016-01-15 09:40:00 SHRT

其中time是：

df.index = df['time'] df.index = pd.to_datetime(df.index)

我想跳过不需要的值（'Plmt'，'Check'，'nan'），创建新列df ['Diff']，'Open' 'SHRT'之间的分钟数之差。 / p>

我正在尝试这样：

df['Status'][df['Status'] == 'SHRT'] - df['Status'][df['Status'] == 'Open']

但在输出中接收NaN值：

time 2016-01-15 08:55:00 NaN 2016-01-15 09:40:00 NaN 2016-01-18 08:30:00 NaN 2016-01-19 14:30:00 NaN 2016-01-19 14:35:00 NaN 2016-01-20 11:10:00 NaN 2016-01-20 11:45:00 NaN

预期的输出必须看起来像： time Status Diff 2016-01-15 08:55:00 Open NaN 2016-01-15 09:40:00 SHRT 00:45:00 2016-02-15 10:00:00 Open NaN 2016-02-15 14:15:00 SHRT 02:15:00

如何获得时间上的差异，任何人都可以帮忙吗？

Answer 1

使用：

#changed data samples for better sample data 
print (df)
                 time Status
0 2016-01-15 08:55:00   Open
1 2016-01-15 09:00:00   Plmt
2 2016-01-15 09:05:00   SHRT
3 2016-01-15 09:10:00   Plmt
4 2016-01-15 09:15:00   Open
5 2016-01-15 09:20:00   Plmt
6 2016-01-15 09:25:00   SHRT
7 2016-01-15 09:30:00   SHRT
8 2016-01-15 09:35:00   Plmt
9 2016-01-15 09:40:00   SHRT

#filter only Open and SHRT
df1 = df[df['Status'].isin(['Open','SHRT'])].copy()
#convert column to datetimes
df1['time'] = pd.to_datetime(df1['time'])
print (df1)
                 time Status
0 2016-01-15 08:55:00   Open
2 2016-01-15 09:05:00   SHRT
4 2016-01-15 09:15:00   Open
6 2016-01-15 09:25:00   SHRT
7 2016-01-15 09:30:00   SHRT
9 2016-01-15 09:40:00   SHRT

#filter only rows with Open and next row SHRT
m1 = (df1['Status'] == 'Open') & (df1['Status'].shift(-1) == 'SHRT')
m2 = (df1['Status'].shift() == 'Open') & (df1['Status'] == 'SHRT')
df2 = df1[m1 | m2].copy()

#create difference column and set NaT by condition
df2['Diff'] = df2['time'].diff().mask(df2['Status'] == 'Open') 
print (df2)
                 time Status     Diff
0 2016-01-15 08:55:00   Open      NaT
2 2016-01-15 09:05:00   SHRT 00:10:00
4 2016-01-15 09:15:00   Open      NaT
6 2016-01-15 09:25:00   SHRT 00:10:00

按列计算时差熊猫

1 个答案: