问题
I have a column of a dataframe that is like this
time
0 2017-03-01 15:30:00
1 2017-03-01 16:00:00
2 2017-03-01 16:30:00
3 2017-03-01 17:00:00
4 2017-03-01 17:30:00
5 2017-03-01 18:00:00
6 2017-03-01 18:30:00
7 2017-03-01 19:00:00
8 2017-03-01 19:30:00
9 2017-03-01 20:00:00
10 2017-03-01 20:30:00
11 2017-03-01 21:00:00
12 2017-03-01 21:30:00
13 2017-03-01 22:00:00
.
.
.
I want to "encode" the time of the day. I want to do this by firsly assigning each half an-hour a integer number. Starting from
00:30:00 --> 1
01:00:00 --> 2
01:30:00 --> 3
02:00:00 --> 4
02:30:00 --> 5
and so on. Therefore we would have 48 numbers (since there are 24 hours). I would like to find the fastest way of transforming my column into a list/column containing those values.
So far I can do this for one value. For instance
2*int(timeDF.iloc[0][11:13]) + int(int(timeDF.iloc[0][14:16])/30)
would transform 15:30:00 into 31.
I think I could do this by doing a loop where instead of using 0 I use an index that loops through the length of the column. However is there a faster way?
one hot encoding
After finding those values, I would use some one-hot-encoder, I think sklearn has one. But the most difficult part is this
stupid solution
labels = []
for date in time:
labels.append(2*int(date[11:13]) + int(int(date[14:16])/30))
This would contain the values and then one could do something like here
回答1:
I think you need map with get_dummies.
Also it seems for first time 0:00 need 0, 0:30 - 1 so using range(48)
#convert to datetimes if necessary
df['time'] = pd.to_datetime(df['time'])
#create dictionary for map
a = dict(zip(pd.date_range('2010-01-01', '2010-01-01 23:59:39', freq='30T').time, range(48)))
#convert time column to times and map by dict
df['a'] = df['time'].dt.time.map(a)
print (df)
time a
0 2017-03-01 15:30:00 31
1 2017-03-01 16:00:00 32
2 2017-03-01 16:30:00 33
3 2017-03-01 17:00:00 34
4 2017-03-01 17:30:00 35
5 2017-03-01 18:00:00 36
6 2017-03-01 18:30:00 37
7 2017-03-01 19:00:00 38
8 2017-03-01 19:30:00 39
9 2017-03-01 20:00:00 40
10 2017-03-01 20:30:00 41
11 2017-03-01 21:00:00 42
12 2017-03-01 21:30:00 43
13 2017-03-01 22:00:00 44
#for one hot encoding use get_dummies
df1 = pd.get_dummies(df['time'].dt.time.map(a))
print (df1)
31 32 33 34 35 36 37 38 39 40 41 42 43 44
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 1 0 0 0 0 0 0 0 0 0 0 0
3 0 0 0 1 0 0 0 0 0 0 0 0 0 0
4 0 0 0 0 1 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 1 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 1 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 1 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 1 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 1 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 1 0 0 0
11 0 0 0 0 0 0 0 0 0 0 0 1 0 0
12 0 0 0 0 0 0 0 0 0 0 0 0 1 0
13 0 0 0 0 0 0 0 0 0 0 0 0 0 1
EDIT:
df1 = pd.get_dummies(df['time'].dt.time.map(a)).reindex(columns=range(48), fill_value=0)
0 1 2 3 4 5 6 7 8 9 ... 38 39 40 41 42 43 44 \
0 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0 0 ... 1 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0 0 ... 0 1 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0 ... 0 0 1 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 1 0 0 0
11 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 1 0 0
12 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 1 0
13 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 1
45 46 47
0 0 0 0
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
6 0 0 0
7 0 0 0
8 0 0 0
9 0 0 0
10 0 0 0
11 0 0 0
12 0 0 0
13 0 0 0
[14 rows x 48 columns]
回答2:
I think this is what you are looking for i.e
x =pd.date_range("00:30", "23:30", freq="30min",format="%HH:%MM").astype(str).str[-8:]
maps = dict(zip(x,np.arange(1,48)))
df['new'] = df['time'].astype(str).str[-8:].map(maps)
pd.get_dummies(df['new']).set_index(df['time'])
Output:
31 32 33 34 35 36 37 38 39 40 41 42 43 44
time
2017-03-01 15:30:00 1 0 0 0 0 0 0 0 0 0 0 0 0 0
2017-03-01 16:00:00 0 1 0 0 0 0 0 0 0 0 0 0 0 0
2017-03-01 16:30:00 0 0 1 0 0 0 0 0 0 0 0 0 0 0
2017-03-01 17:00:00 0 0 0 1 0 0 0 0 0 0 0 0 0 0
2017-03-01 17:30:00 0 0 0 0 1 0 0 0 0 0 0 0 0 0
2017-03-01 18:00:00 0 0 0 0 0 1 0 0 0 0 0 0 0 0
2017-03-01 18:30:00 0 0 0 0 0 0 1 0 0 0 0 0 0 0
2017-03-01 19:00:00 0 0 0 0 0 0 0 1 0 0 0 0 0 0
2017-03-01 19:30:00 0 0 0 0 0 0 0 0 1 0 0 0 0 0
2017-03-01 20:00:00 0 0 0 0 0 0 0 0 0 1 0 0 0 0
2017-03-01 20:30:00 0 0 0 0 0 0 0 0 0 0 1 0 0 0
2017-03-01 21:00:00 0 0 0 0 0 0 0 0 0 0 0 1 0 0
2017-03-01 21:30:00 0 0 0 0 0 0 0 0 0 0 0 0 1 0
2017-03-01 22:00:00 0 0 0 0 0 0 0 0 0 0 0 0 0 1
来源：https://stackoverflow.com/questions/46607306/python-numpy-and-pandas-transforming-timestamp-data-into-one-hot-encoding