Python:Numpy和Pandas将时间戳/数据转换为一个热编码
我有一列数据框,它是这样的Python:Numpy和Pandas将时间戳/数据转换为一个热编码,python,pandas,date,numpy,encoding,Python,Pandas,Date,Numpy,Encoding,我有一列数据框,它是这样的 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
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
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.
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我想“编码”一天中的时间。我想先给每半小时分配一个整数。从
00:30:00 --> 1
01:00:00 --> 2
01:30:00 --> 3
02:00:00 --> 4
02:30:00 --> 5
等等。因此,我们将有48个数字(因为有24小时)。我想找到将我的列转换为包含这些值的列表/列的最快方法
到目前为止,我只需要一个值就可以做到这一点。比如说
2*int(timeDF.iloc[0][11:13])+int(int(timeDF.iloc[0][14:16])/30)
将15:30:00
转换为31
我想我可以通过做一个循环来做到这一点,在这个循环中,我不用0
而是使用一个循环遍历列长度的索引。然而,有没有更快的方法
一个热编码
找到这些值后,我会使用某个热编码器,我认为sklearn有一个。但最困难的是这一点
愚蠢的解决方案
labels = []
for date in time:
labels.append(2*int(date[11:13]) + int(int(date[14:16])/30))
这将包含这些值,然后可以执行类似于我认为您需要的操作
而且似乎第一次0:00
需要0
,0:30
-1
,所以使用范围(48)
编辑:
我想这就是你要找的
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'])
输出:
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
31 32 33 34 35 36 37 38 39 40 41 42 43 44
时间
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
您想从获取的列表中将值映射到数据帧吗?是的,如果我正确理解您的问题。数据帧有一列名为
time
。此列中有日期和时间。我只关心一天中的时间。我想在回归中使用它作为一个特性。因此,我需要将时间映射到某个数字。我想这样做的方式是,你拿这个列,把时间转换/映射到相应的数字(例如00:30:00
到1
),然后把这个数字转换成一个热编码器,例如[1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
,谢谢!那么,您将如何进行热编码呢?例如:[1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
对于00:30:00,即1
我的解决方案是什么?等等,我不明白你的解决方案的最后一部分。。为什么“31”(对应于“15:30:00”)被编码为“[1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0”,而1[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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]
然后需要按范围重新索引列,请参见最后一次编辑。您的实际数据是更多列,如48
?但通过这种方式,“31”被映射到第一个条目中带有“1”的列表,而不是“30”。这是基于您提供的示例数据。如果您有从0:00开始的数据,则它将自动更新
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]
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'])
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