series

I have a data frame called data , which has a column Dates like this, Dates 0 2015-05-13 23:53:00 1 2015-05-13 23:53:00 2 2015-05-13 23:33:00 3 2015-05-13 23:30:00 4 2015-05-13 23:30:00 I know how to add a column to data frame, but how to divide Dates to Day Time 0 2015-05-13 23:53:00 1 2015-05-13 23:53:00 2 2015-05-13 23:33:00 3 2015-05-13 23:30:00 4 2015-05-13 23:30:00 If your series is s , then this will create such a DataFrame: pd.DataFrame({ 'date': pd.to_datetime(s).dt.date, 'time': pd.to_datetime(s).dt.time}) as once you convert the series using pd.to_datetime , then the dt member can

I want to make bar chart using jfreechart such that the bars which belong to the same category should be displayed adjacent without any gaps. The categories should be displayed with gaps. Also each category may have different number of bars. How it can be achived using Jfreechart library? Following image is the sample of what I require. Here all the bars of same category should be of same color and with no gap(or a very little gap). Thanks in advance, Abhinav I am aware of the age of this post. Anyway I am posting my solution, maybe someone else who will find himself here looking for the

In my script I have df['Time'] as shown below. 497 2017-08-06 11:00:00 548 2017-08-08 15:00:00 580 2017-08-10 04:00:00 646 2017-08-12 23:00:00 Name: Time, dtype: datetime64[ns] But when i do t1=pd.Timestamp(df['Time'][0]) I get an error like this : KeyError: 0 Do I need any type conversion here, if yes then how it can be fixed? cᴏʟᴅsᴘᴇᴇᴅ You're looking for df.iloc . df['Time'].iloc[0] df['Time'][0] would've worked if your series had an index beginning from 0 And if need scalar only use Series.iat : df['Time'].iat[0] 来源： https://stackoverflow.com/questions/46153647/keyerror-0-when-accessing

Suppose I have some number a , and I want to get vector [ 1 , a , a^2 , ... , a^N ] . I use [ 1 , cumprod( a * ones( 1 , N - 1 ) ) ] code. What is the best (and propably efficient) way to do it? What about a.^[0:N] ? ThibThib's answer is absolutely correct, but it doesn't generalize very easily if a happens to a vector. So as a starting point: > a= 2 a = 2 > n= 3 n = 3 > a.^[0: n] ans = 1 2 4 8 Now you could also utilize the built-in function vander (although the order is different, but that's easily fixed if needed), to produce: > vander(a, n+ 1) ans = 8 4 2 1 And with vector valued a : > a=

I have a data frame which has a column usage_duration (which is the difference of two another columns in datetime format). It looks like below: processid, userid, usage_duration 17613,root,0 days 23:41:03.000000000 17641,root,2 days 04:05:26.000000000 13848,acs,0 days 00:00:50.000000000 3912,acs,0 days 06:07:38.000000000 6156,acs,0 days 17:22:43.000000000 Now I wanted to convert the same into minutes. It should look like as below: processid, userid, usage_duration_min 17613,root,1421 17641,root,3125 13848,acs,0 3912,acs,367 6156,acs,1042 Can someone let me know how is it possible? Highly

how can I access to the next point of the series from the tooltip formatter. Because I want to do a sum between both points. Like this.y + next.y. But I don't know how to have an access to the next point. If someone have an answer. Thanks This need to be done in a few steps: get x-index according to x-value: var index = this.series.xData.indexOf(this.x); now get y-value: var nextY = this.series.yData[index+1]; And all you need to do is to sum values, like this: var sum = this.y + nextY; . 来源： https://stackoverflow.com/questions/21578555/how-to-get-next-point-in-highcharts-tooltip