gpt4 book ai didi

python - 从不同大小的元组填充 DataFrame

转载 作者:太空宇宙 更新时间:2023-11-03 16:10:04 25 4
gpt4 key购买 nike

我有全天的数据。我对它进行了聚类,然后计算了每小时每个聚类的比率(权重)(并非所有聚类都存在于所有小时中)。(数据帧 time_df )

   cluster                             Date
0 1 2014-02-28 14:24:59.535000+02:00
1 1 2014-02-28 14:26:35.019000+02:00
2 1 2014-02-28 14:27:37.213000+02:00
3 2 2014-02-28 14:28:35.246000+02:00
4 2 2014-02-28 14:29:37.283000+02:00

我按小时分组,并使用 np bincount 计算每个簇的权重:

group_by_hour = time_df.groupby(time_df.Date.dt.hour)
cluster_ids_hour = group_by_hour.cluster.\
apply(lambda arr: list(range(0,(arr+1).max()+1)))
cluster_ratio_hour = group_by_hour.cluster.\
apply(lambda arr: 1.0*np.bincount(arr+1)/len(arr))

这给出了每小时不同的簇数组大小及其权重它尝试构建一个数据框

pd.DataFrame(temp, columns=['小时','集群','比率'])

但是我得到了以下信息:

   hour   clusters                                           weights
0 14 [0] [1.0]
1 15 [0, 1] [0.488888888889, 0.511111111111]
2 16 [0, 1, 2] [0.302325581395, 0.162790697674, 0.53488372093]
3 17 [0, 1, 2] [0.0, 0.0, 1.0]
4 18 [0, 1, 2] [0.0, 0.0, 1.0]
5 19 [0, 1, 2] [0.0, 0.0, 1.0]
6 20 [0, 1, 2] [0.0, 0.0, 1.0]
7 21 [0, 1, 2] [0.0, 0.0, 1.0]
8 22 [0, 1, 2] [0.0, 0.0, 1.0]
9 23 [0, 1, 2] [0.0, 0.0, 1.0]

如何才能将集群作为索引,将时间作为列?

    0    1    2    3    4    ...
0 0.2 0.6 0.4 0.0 0.6
1 0.0 0.4 0.1 0.0 0.4
2 0.8 0.0 0.5 1.0 0.0

最佳答案

我认为你可以使用:

import pandas as pd
import numpy as np

time_df = pd.DataFrame({'cluster': {0: 1, 1: 1, 2: 1, 3: 2, 4: 2, 5: 1, 6: 1, 7: 2},
'Date': {0: pd.Timestamp('2014-02-28 12:24:59.535000'),
1: pd.Timestamp('2014-02-28 12:26:35.019000'),
2: pd.Timestamp('2014-02-28 12:27:37.213000'),
3: pd.Timestamp('2014-02-28 12:28:35.246000'),
4: pd.Timestamp('2014-02-28 12:29:37.283000'),
5: pd.Timestamp('2014-02-28 13:27:37.213000'),
6: pd.Timestamp('2014-02-28 14:28:35.246000'),
7: pd.Timestamp('2014-02-28 14:29:37.283000')}})

print (time_df)
Date cluster
0 2014-02-28 12:24:59.535 1
1 2014-02-28 12:26:35.019 1
2 2014-02-28 12:27:37.213 1
3 2014-02-28 12:28:35.246 2
4 2014-02-28 12:29:37.283 2
5 2014-02-28 13:27:37.213 1
6 2014-02-28 14:28:35.246 1
7 2014-02-28 14:29:37.283 2
group_by_hour = time_df.groupby(time_df.Date.dt.hour)
cluster_ids_hour = group_by_hour.cluster.\
apply(lambda arr: list(range(0,(arr+1).max()+1)))
cluster_ratio_hour = group_by_hour.cluster.\
apply(lambda arr: 1.0*np.bincount(arr+1)/len(arr))

print (cluster_ids_hour)
Date
12 [0, 1, 2, 3]
13 [0, 1, 2]
14 [0, 1, 2, 3]
Name: cluster, dtype: object

print (cluster_ratio_hour)
Date
12 [0.0, 0.0, 0.6, 0.4]
13 [0.0, 0.0, 1.0]
14 [0.0, 0.0, 0.5, 0.5]
Name: cluster, dtype: object

#create DataFrames from both columns and concate them
df1 = pd.DataFrame(cluster_ids_hour.values.tolist(), index=cluster_ids_hour.index)
#print (df1)

df2 = pd.DataFrame(cluster_ratio_hour.values.tolist(), index=cluster_ratio_hour.index)
#print (df2)
df = pd.concat([df1, df2], axis=1, keys=('clusters','weights'))
print (df)
clusters weights
0 1 2 3 0 1 2 3
Date
12 0 1 2 3.0 0.0 0.0 0.6 0.4
13 0 1 2 NaN 0.0 0.0 1.0 NaN
14 0 1 2 3.0 0.0 0.0 0.5 0.5
#reshape, cast clusters column to integer    
df = df.stack().reset_index(level=1, drop=True).reset_index()
df['clusters'] = df['clusters'].astype(int)
#pivoting, fill NaN by 0
df = df.pivot(index='clusters', columns='Date', values='weights').fillna(0)

df.index.name = None
df.columns.name = None
print (df)
12 13 14
0 0.0 0.0 0.0
1 0.0 0.0 0.0
2 0.6 1.0 0.5
3 0.4 0.0 0.5

关于python - 从不同大小的元组填充 DataFrame,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39409544/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com