在python中按时间分组和绘制数据

Question

我有一个csv文件，我试图绘制每月一些值的平均值。我的csv文件的结构如下所示，所以我认为我应该每天对数据进行分组，然后按月分组，以便计算平均值。

bean

我正在使用此代码：

timestamp,heure,lat,lon,impact,type
2007-01-01 00:00:00,13:58:43,33.837,-9.205,10.3,1
2007-01-02 00:00:00,00:07:28,34.5293,-10.2384,17.7,1
2007-01-02 00:00:00,23:01:03,35.0617,-1.435,-17.1,2
2007-01-03 00:00:00,01:14:29,36.5685,0.9043,36.8,1
2007-01-03 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1

但是，我一直遇到这样的错误：

  

KeyError：＆＃39;找不到石斑鱼名称时间戳＆＃39;

任何想法??

Answer 1

您收到此错误是因为您已将timestamp列设置为index。尝试从key='timestamp'或TimeGrouper()方法移除set_index，并按预期进行分组：

daily = df.set_index('timestamp').groupby(pd.TimeGrouper(freq='D', axis=1), axis=1)['impact'].count()

或

daily = df.groupby(pd.TimeGrouper(key='timestamp', freq='D', axis=1), axis=1)['impact'].count()

Answer 2

我相信你需要DataFrame.resample。

还需要通过read_csv中的参数timestamp和DataTimeindex将parse_dates转换为index_col。

names =["timestamp","heure","lat","lon","impact","type"]
data = pd.read_csv('fou.txt',names=names, parse_dates=['timestamp'],index_col=['timestamp'])
print (data.head())

#your code
daily = data.groupby(pd.TimeGrouper(freq='D'))['impact'].count()
monthly = daily.groupby(pd.TimeGrouper(freq='M')).mean()
ax = monthly.plot(kind='bar')
plt.show()

#more simpliest
daily = data.resample('D')['impact'].count()
monthly = daily.resample('M').mean()
ax = monthly.plot(kind='bar')
plt.show()

同时检查是否真的需要count，而不是size。 What is the difference between size and count in pandas?

daily = data.resample('D')['impact'].size()
monthly = daily.resample('M').mean()
ax = monthly.plot(kind='bar')
plt.show()