I'm currently working with some data that I receive from an engineering plant, the data comes out(roughly) as the following :
df = pd.DataFrame({'ID' : np.random.randint(1,25,size=5),
'on/off' : np.random.randint(0,2,size=5),
'Time' : pd.date_range(start='01/01/2019',periods=5,freq='5s')})
print(df)
ID on/off Time
0 17 0 2019-01-01 00:00:00
1 21 0 2019-01-01 00:00:05
2 12 1 2019-01-01 00:00:10
3 12 1 2019-01-01 00:00:15
4 12 0 2019-01-01 00:00:20
the 0 and 1 in the on/off column correspond to when a machine is on or off (0 = on 1 = off)
currently, I use the following line of beautiful code to get the difference between my column as the data is rolling
df['Time Difference'] = (df.time - df.time.shift())
print(df)
ID on/off Time Time Difference
0 17 0 2019-01-01 00:00:00 NaT
1 21 0 2019-01-01 00:00:05 00:00:05
2 12 1 2019-01-01 00:00:10 00:00:05
3 12 1 2019-01-01 00:00:15 00:00:05
4 12 0 2019-01-01 00:00:20 00:00:05
now as this dataframe is quite verbose (each week I'll receive about 150k rows)
what would be the best way to sum amount time a machine is off (where df['on/off] == 1) until the next 0 comes along? so in the above example for the 1st of January 2019 the machine of ID 12 didn't run for 15 seconds until it resumed at 00:00:20
Here's an approach that works for a simple example of one machine that varies between on and off during the course of one day. It works regardless of whether the machine is in on or off state in the first row.
df = pd.DataFrame({'ID': [12, 12, 12, 12, 12],
'on/off': [0,0,1,0,1],
'Time': ['2019-01-01 00:00:00', '2019-01-01 00:00:05', '2019-01-01 00:00:10','2019-01-01 00:00:15','2019-01-01 00:00:20']
})
ID on/off Time
0 12 0 2019-01-01 00:00:00
1 12 0 2019-01-01 00:00:05
2 12 1 2019-01-01 00:00:10
3 12 0 2019-01-01 00:00:15
4 12 1 2019-01-01 00:00:20
First I made sure the Time column dtype is datetime64:
df['Time'] = pd.to_datetime(df['Time'])
Then I get the indices of all rows where the state changed (either from off to on, or from on to off:
s = df[df['on/off'].shift(1) != df['on/off']].index
df = df.loc[s]
Then I create a column called time shift, which shows the timestamp of the most recent row where power state changed:
df['time shift'] = df['Time'].shift(1)
At this point the dataframe looks like this:
ID on/off Time time shift
0 12 0 2019-01-01 00:00:00 NaT
2 12 1 2019-01-01 00:00:10 2019-01-01 00:00:00
3 12 0 2019-01-01 00:00:15 2019-01-01 00:00:10
4 12 1 2019-01-01 00:00:20 2019-01-01 00:00:15
Now, since we want to count the duration that the machine was off, I look at only the row indices where the state became on:
r = df[df['on/off'] == 1].index
df = df.loc[r]
At this point, the dataframe looks as it does below. Notice that the time shift column is displaying the point at which the machine most recently turned off, prior to the time being displayed in Time column, which is the timestamp when the machine turned back on. Finding the difference between these two columns will give us the length of each duration that the machine was off during the day:
ID on/off Time time shift
2 12 1 2019-01-01 00:00:10 2019-01-01 00:00:00
4 12 1 2019-01-01 00:00:20 2019-01-01 00:00:15
The following line calculates total off-time, by summing the durations of each period that the machine was in its off state:
(df['Time'] - df['time shift']).sum()
Which outputs:
Timedelta('0 days 00:00:15')
Some additional context on how the Pandas .shift() method works:
Shift takes all the rows in a column, and moves them either forward or back by a certain amount. .shift(1) tells pandas to move the index of each row forward, or up, by 1. .shift(-1) tells pandas to move the index of each row back, or down, by 1. Alternately put, .shift(1) lets you look at the value of a column at the previous row index, and .shift(-1) lets you look at the value of a column at the next row index, relative a given row in a column. It's a handy way to compare a column's values across different rows, without resorting to for-loops.
Related
I have the following data frame where the column hour shows hours of the day in int64 form. I'm trying to convert that into a time format; so that hour 1 would show up as '01:00'. I then want to add this to the date column and convert it into a timestamp index.
Using the datetime function in pandas resulted in the column "hr2", which is not what I need. I'm not sure I can even apply datetime directly, as the original data (i.e. in column "hr") is not really a date time format to begin with. Google searches so far have been unproductive.
While I am still in the dark concerning the format of your date column. I will assume the Date column is a string object and the hr column is an int64 object. To create the column TimeStamp in pandas tmestamp format this is how I would proceed>
Given df:
Date Hr
0 12/01/2010 1
1 12/01/2010 2
2 12/01/2010 3
3 12/01/2010 4
4 12/02/2010 1
5 12/02/2010 2
6 12/02/2010 3
7 12/02/2010 4
df['TimeStamp'] = df.apply(lambda row: pd.to_datetime(row['Date']) + pd.to_timedelta(row['Hr'], unit='H'), axis = 1)
yields:
Date Hr TimeStamp
0 12/01/2010 1 2010-12-01 01:00:00
1 12/01/2010 2 2010-12-01 02:00:00
2 12/01/2010 3 2010-12-01 03:00:00
3 12/01/2010 4 2010-12-01 04:00:00
4 12/02/2010 1 2010-12-02 01:00:00
5 12/02/2010 2 2010-12-02 02:00:00
6 12/02/2010 3 2010-12-02 03:00:00
7 12/02/2010 4 2010-12-02 04:00:00
The timestamp column can then be used as your index.
I'm looking to understand the number of times we are in an 'Abnormal State' before we have an 'Event'. My objective is to modify my dataframe to get the following output where everytime we reach an 'event', the 'Abnormal State Grouping' resets to count from 0.
We can go through a number of 'Abnormal States' before we reach an 'Event', which is deemed a failure. (i.e. The lightbulb is switched on and off for several periods before it finally shorts out resulting in an event).
I've written the following code to get my AbnormalStateGroupings to increment into relevant groupings for my analysis which has worked fine. However, we want to 'reset' the count of our 'AbnormalStates' after each event (i.e. lightbulb failure):
dataframe['AbnormalStateGrouping'] = (dataframe['AbnormalState']!=dataframe['AbnormalState'].shift()).cumsum()
I have created an additional column which let's me know what 'event' we are at via:
dataframe['Event_Or_Not'].cumsum() #I have a boolean representation of the Event Column represented and we use .cumsum() to get the relevant groupings (i.e. 1st Event, 2nd Event, 3rd Event etc.)
I've come close previously using the following:
eventOrNot = dataframe['Event'].eq(0)
eventMask = (eventOrNot.ne(eventOrNot.shift())&eventOrNot).cumsum()
dataframe['AbnormalStatePerEvent'] =dataframe.groupby(['Event',eventMask]).cumcount().add(1)
However, this hasn't given me the desired output that I'm after (as per below).
I think I'm close however - Could anyone please advise what I could try to do next so that for each lightbulb failure, the abnormal state count resets and starts counting the # of abnormal states we have gone through before the next lightbulb failure?
State I want to get to with AbnormalStateGrouping
You would note that when an 'Event' is detected, the Abnormal State count resets to 1 and then starts counting again.
Current State of Dataframe
Please find an attached data source below:
https://filebin.net/ctjwk7p3gulmbgkn
I assume that your source DataFrame has only Date/Time (either string
or datetime), Event (string) and AbnormalState (int) columns.
To compute your grouping column, run:
dataframe['AbnormalStateGrouping'] = dataframe.groupby(
dataframe['Event'][::-1].notnull().cumsum()).AbnormalState\
.apply(lambda grp: (grp != grp.shift()).cumsum())
The result, for your initial source data, included as a picture, is:
Date/Time Event AbnormalState AbnormalStateGrouping
0 2018-01-01 01:00 NaN 0 1
1 2018-01-01 02:00 NaN 0 1
2 2018-01-01 03:00 NaN 1 2
3 2018-01-01 04:00 NaN 1 2
4 2018-01-01 05:00 NaN 0 3
5 2018-01-01 06:00 NaN 0 3
6 2018-01-01 07:00 NaN 0 3
7 2018-01-01 08:00 NaN 1 4
8 2018-01-01 09:00 NaN 1 4
9 2018-01-01 10:00 NaN 0 5
10 2018-01-01 11:00 NaN 0 5
11 2018-01-01 12:00 NaN 0 5
12 2018-01-01 13:00 NaN 1 6
13 2018-01-01 14:00 NaN 1 6
14 2018-01-01 15:00 NaN 0 7
15 2018-01-01 16:00 Event 0 7
16 2018-01-01 17:00 NaN 1 1
17 2018-01-01 18:00 NaN 1 1
18 2018-01-01 19:00 NaN 0 2
19 2018-01-01 20:00 NaN 0 2
Note the way of grouping:
dataframe['Event'][::-1].notnull().cumsum()
Due to [::-1], cumsum function is computed from the last row
to the first.
Thus:
rows with hours 01:00 thru 16:00 are in group 1,
remaining rows (hour 17:00 thru 20:00) are in group 0.
Then, to AbnormalState, separately for each group, a lambda function
is applied, so each cumulative sum starts from 1 just in each group
(after each Event).
Edit following the comment as of 22:18:12Z
The reason why I compute the cumsum for grouping in reversed order
is that when you run it in normal order:
dataframe['Event'].notnull().cumsum()
then:
rows with index 0 thru 14 (before the row with Event) have
this sum == 0,
row with index 15 and following rows have this sum == 1.
Try yourself both versions, without and with [::-1].
The result in normal order (without [::-1]) is that:
Event row is in the same group with the following rows,
so the reset occurs just on this row.
To check the whole result, run my code without [::-1] and you will see
that the ending part of the result contains:
Date/Time Event AbnormalState AbnormalStateGrouping
14 2018-01-01 15:00:00 NaN 0 7
15 2018-01-01 16:00:00 Event 0 1
16 2018-01-01 17:00:00 NaN 1 2
17 2018-01-01 18:00:00 NaN 1 2
18 2018-01-01 19:00:00 NaN 0 3
19 2018-01-01 20:00:00 NaN 0 3
so that the Event row to has AbnormalStateGrouping == 1.
But you want this row to have AbnormalStateGrouping in a sequence of
previous grouping states (in this case 7) and reset should occur
from the next row on.
So the Event row should be in same group with preceding rows, what
is the result of my code.
I'm trying to count a frequency of 2 events by the month using 2 columns from my df. What I have done so far has counted all events by the unique time which is not efficient enough as there are too many results. I wish to create a graph with the results afterwards.
I've tried adapting my code by the answers on the SO questions:
[How to groupby time series by 10 minutes using pandas?
[Counting frequency of occurrence by month-year using python panda
[Pandas Groupby using time frequency
but can not seem to get the command working when I input freq='day' within the groupby command.
My code is:
print(df.groupby(['Priority', 'Create Time']).Priority.count())
which initially produced something like 170000 results in the structure of the following:
Priority Create Time
1.0 2011-01-01 00:00:00 1
2011-01-01 00:01:11 1
2011-01-01 00:02:10 1
...
2.0 2011-01-01 00:01:25 1
2011-01-01 00:01:35 1
...
But now for some reason (I'm using Jupyter Notebook) it only produces:
Priority Create Time
1.0 2011-01-01 00:00:00 1
2011-01-01 00:01:11 1
2011-01-01 00:02:10 1
2.0 2011-01-01 00:01:25 1
2011-01-01 00:01:35 1
Name: Priority, dtype: int64
No idea why the output has changed to only 5 results (maybe I unknowingly changed something).
I would like the results to be in the following format:
Priority month Count
1.0 2011-01 a
2011-02 b
2011-03 c
...
2.0 2011-01 x
2011-02 y
2011-03 z
...
Top points for showing how to change the frequency correctly for other values as well, for example hour/day/month/year. With the answers please could you explain what is going on in your code as I am new and learning pandas and wish to understand the process. Thank you.
One possible solution is convert datetime column to months periods by Series.dt.to_period:
print(df.groupby(['Priority', df['Create Time'].dt.to_period('m')]).Priority.count())
Or use Grouper:
print(df.groupby(['Priority', pd.Grouper(key='Create Time', freq='MS')]).Priority.count())
Sample:
np.random.seed(123)
df = pd.DataFrame({'Create Time':pd.date_range('2019-01-01', freq='10D', periods=10),
'Priority':np.random.choice([0,1], size=10)})
print (df)
Create Time Priority
0 2019-01-01 0
1 2019-01-11 1
2 2019-01-21 0
3 2019-01-31 0
4 2019-02-10 0
5 2019-02-20 0
6 2019-03-02 0
7 2019-03-12 1
8 2019-03-22 1
9 2019-04-01 0
print(df.groupby(['Priority', df['Create Time'].dt.to_period('m')]).Priority.count())
Priority Create Time
0 2019-01 3
2019-02 2
2019-03 1
2019-04 1
1 2019-01 1
2019-03 2
Name: Priority, dtype: int64
print(df.groupby(['Priority', pd.Grouper(key='Create Time', freq='MS')]).Priority.count())
Priority Create Time
0 2019-01-01 3
2019-02-01 2
2019-03-01 1
2019-04-01 1
1 2019-01-01 1
2019-03-01 2
Name: Priority, dtype: int64
I have been using datetimes from the datetime library in Python, and making them timezone aware with pytz. I have then been using them as dates in Pandas DataFrames and trying to use Pandas's apply function and the ".day", ".hour", ".minute" etc. methods of the datetimes to create columns with just the day, hour, or minute. Surprisingly, it gives the UTC values. Is there a way to return the local day, hour, or minute? Simply adding the offset is not good enough, because the offset to UTC changes with daylight savings time.
Many thanks!
Here is an example of what I am talking about:
import pandas as pd
import datetime as dt
import pytz
# Simply return the hour of a date
def get_hour(dt1):
return dt1.hour
# Create a date column to segment by month
# Create the date list
PST = pytz.timezone('US/Pacific')
start = PST.localize(dt.datetime(2016, 1, 1))
actuals_dates = [start + dt.timedelta(hours=x) for x in range(8760)]
# Outside of this context, you can get the hour
print ''
print 'Hour at the start date:'
print get_hour(start)
print ''
#add it to a pandas DataFrame as a column
shapes = pd.DataFrame()
shapes['actuals dates'] = actuals_dates
# create a column for the hour
shapes['actuals hour'] = shapes['actuals dates'].apply(get_hour)
# Print the first 24 hours
print shapes.head(24)
Will return:
Hour at the start date:
0
actuals dates actuals hour
0 2016-01-01 00:00:00-08:00 8
1 2016-01-01 01:00:00-08:00 9
2 2016-01-01 02:00:00-08:00 10
3 2016-01-01 03:00:00-08:00 11
4 2016-01-01 04:00:00-08:00 12
5 2016-01-01 05:00:00-08:00 13
6 2016-01-01 06:00:00-08:00 14
7 2016-01-01 07:00:00-08:00 15
8 2016-01-01 08:00:00-08:00 16
9 2016-01-01 09:00:00-08:00 17
10 2016-01-01 10:00:00-08:00 18
11 2016-01-01 11:00:00-08:00 19
12 2016-01-01 12:00:00-08:00 20
13 2016-01-01 13:00:00-08:00 21
14 2016-01-01 14:00:00-08:00 22
15 2016-01-01 15:00:00-08:00 23
16 2016-01-01 16:00:00-08:00 0
17 2016-01-01 17:00:00-08:00 1
18 2016-01-01 18:00:00-08:00 2
19 2016-01-01 19:00:00-08:00 3
20 2016-01-01 20:00:00-08:00 4
21 2016-01-01 21:00:00-08:00 5
22 2016-01-01 22:00:00-08:00 6
23 2016-01-01 23:00:00-08:00 7
Using a list comprehension seems to do the trick:
shapes['hour'] = [ts.hour for ts in shapes['actuals dates']]
shapes.head()
actuals dates actuals hour hour
0 2016-01-01 00:00:00-08:00 8 0
1 2016-01-01 01:00:00-08:00 9 1
2 2016-01-01 02:00:00-08:00 10 2
3 2016-01-01 03:00:00-08:00 11 3
4 2016-01-01 04:00:00-08:00 12 4
Per the reminder from #Jeff, you can also use the dt accessor functions, e.g.:
>>> shapes['actuals dates'].dt.hour.head()
0 0
1 1
2 2
3 3
4 4
Name: actuals dates, dtype: int64
I have a dataframe indexed using a 12hr frequency datetime:
id mm ls
date
2007-09-27 00:00:00 1 0 0
2007-09-27 12:00:00 1 0 0
2007-09-28 00:00:00 1 15 0
2007-09-28 12:00:00 NaN NaN 0
2007-09-29 00:00:00 NaN NaN 0
Timestamp('2007-09-27 00:00:00', offset='12H')
I use column 'ls' as a binary variable with default value '0' using:
data['ls'] = 0
I have a list of days in the form '2007-09-28' from which I wish to update all 'ls' values from 0 to 1.
id mm ls
date
2007-09-27 00:00:00 1 0 0
2007-09-27 12:00:00 1 0 0
2007-09-28 00:00:00 1 15 1
2007-09-28 12:00:00 NaN NaN 1
2007-09-29 00:00:00 NaN NaN 0
Timestamp('2007-09-27 00:00:00', offset='12H')
I understand how this can be done using another column variable ie:
data.ix[data.id == '1'], ['ls'] = 1
yet this does not work using datetime index.
Could you let me know what the method for datetime index is?
You have a list of days in the form '2007-09-28':
days = ['2007-09-28', ...]
then you can modify your df using:
df['ls'][pd.DatetimeIndex(df.index.date).isin(days)] = 1