I am working with a pandas dataframe where I have the following two columns: "personID" and "points". I would like to create a third variable ("localMin") which will store the minimum value of the column "points" at each point in the dataframe as compared with all previous values in the "points" column for each personID (see image below).
Does anyone have an idea how to achieve this most efficiently? I have approached this problem using shift() with different period sizes, but of course, shift is sensitive to variations in the sequence and doesn't always produce the output I would expect.
Thank you in advance!
Use groupby.cummin:
df['localMin'] = df.groupby('personID')['points'].cummin()
Example:
df = pd.DataFrame({'personID': list('AAAAAABBBBBB'),
'points': [3,4,2,6,1,2,4,3,1,2,6,1]
})
df['localMin'] = df.groupby('personID')['points'].cummin()
output:
personID points localMin
0 A 3 3
1 A 4 3
2 A 2 2
3 A 6 2
4 A 1 1
5 A 2 1
6 B 4 4
7 B 3 3
8 B 1 1
9 B 2 1
10 B 6 1
11 B 1 1
Suppose I have pandas DataFrame like this:
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4], 'value':[1,2,3,1,2,3,4,1,1]})
which looks like:
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
I want to get a new DataFrame with top 2 records for each id, like this:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
I can do it with numbering records within group after groupby:
dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index()
which looks like:
id level_1 index value
0 1 0 0 1
1 1 1 1 2
2 1 2 2 3
3 2 0 3 1
4 2 1 4 2
5 2 2 5 3
6 2 3 6 4
7 3 0 7 1
8 4 0 8 1
then for the desired output:
dfN[dfN['level_1'] <= 1][['id', 'value']]
Output:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).
Did you try
df.groupby('id').head(2)
Output generated:
id value
id
1 0 1 1
1 1 2
2 3 2 1
4 2 2
3 7 3 1
4 8 4 1
(Keep in mind that you might need to order/sort before, depending on your data)
EDIT: As mentioned by the questioner, use
df.groupby('id').head(2).reset_index(drop=True)
to remove the MultiIndex and flatten the results:
id value
0 1 1
1 1 2
2 2 1
3 2 2
4 3 1
5 4 1
Since 0.14.1, you can now do nlargest and nsmallest on a groupby object:
In [23]: df.groupby('id')['value'].nlargest(2)
Out[23]:
id
1 2 3
1 2
2 6 4
5 3
3 7 1
4 8 1
dtype: int64
There's a slight weirdness that you get the original index in there as well, but this might be really useful depending on what your original index was.
If you're not interested in it, you can do .reset_index(level=1, drop=True) to get rid of it altogether.
(Note: From 0.17.1 you'll be able to do this on a DataFrameGroupBy too but for now it only works with Series and SeriesGroupBy.)
Sometimes sorting the whole data ahead is very time consuming.
We can groupby first and doing topk for each group:
g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)
df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))
Here sort values ascending false gives similar to nlargest and True gives similar to nsmallest.
The value inside the head is the same as the value we give inside nlargest to get the number of values to display for each group.
reset_index is optional and not necessary.
This works for duplicated values
If you have duplicated values in top-n values, and want only unique values, you can do like this:
import pandas as pd
ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])
id first_name last_name department salary
24 12 Shandler Bing Audit 110000
25 14 Jason Tom Audit 100000
26 16 Celine Anston Audit 100000
27 15 Michale Jackson Audit 70000
If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:
(df.groupby('department')['salary']
.apply(lambda ser: ser.drop_duplicates().nlargest(3))
.droplevel(level=1)
.sort_index()
.reset_index()
)
This gives
department salary
0 Audit 110000
1 Audit 100000
2 Audit 70000
3 Management 250000
4 Management 200000
5 Management 150000
6 Sales 220000
7 Sales 200000
8 Sales 150000
To get the first N rows of each group, another way is via groupby().nth[:N]. The outcome of this call is the same as groupby().head(N). For example, for the top-2 rows for each id, call:
N = 2
df1 = df.groupby('id', as_index=False).nth[:N]
To get the largest N values of each group, I suggest two approaches.
First sort by "id" and "value" (make sure to sort "id" in ascending order and "value" in descending order by using the ascending parameter appropriately) and then call groupby().nth[].
N = 2
df1 = df.sort_values(by=['id', 'value'], ascending=[True, False])
df1 = df1.groupby('id', as_index=False).nth[:N]
Another approach is to rank the values of each group and filter using these ranks.
# for the entire rows
N = 2
msk = df.groupby('id')['value'].rank(method='first', ascending=False) <= N
df1 = df[msk]
# for specific column rows
df1 = df.loc[msk, 'value']
Both of these are much faster than groupby().apply() and groupby().nlargest() calls as suggested in the other answers on here(1, 2, 3). On a sample with 100k rows and 8000 groups, a %timeit test showed that it was 24-150 times faster than those solutions.
Also, instead of slicing, you can also pass a list/tuple/range to a .nth() call:
df.groupby('id', as_index=False).nth([0,1])
# doesn't even have to be consecutive
# the following returns 1st and 3rd row of each id
df.groupby('id', as_index=False).nth([0,2])
I have a Pandas Dataframe with data about calls. Each call has a unique ID and each customer has an ID (but can have multiple Calls). A third column gives a day. For each customer I want to calculate the maximum number of calls made in a period of 7 days.
I have been using the following code to count the number of calls within 7 days of the call on each row:
df['ContactsIN7Days'] = df.apply(lambda row: len(df[(df['PersonID']==row['PersonID']) & (abs(df['Day'] - row['Day']) <=7)]), axis=1)
Output:
CallID Day PersonID ContactsIN7Days
6 2 3 2
3 14 2 2
1 8 1 1
5 1 3 2
2 12 2 2
7 100 3 1
This works, however this is going to be applied on a big data set. Would there be a way to make this more efficient. Through vectorization?
IIUC this is a convoluted, but I think effective solution to your issue. Note that the order of your dataframe is modified as a result, and that your Day column is modified to a timedelta dtype:
Starting from your dataframe df:
CallID Day PersonID
0 6 2 3
1 3 14 2
2 1 8 1
3 5 1 3
4 2 12 2
5 7 100 3
Start by modifying Day to a timedelta series:
df['Day'] = pd.to_timedelta(df['Day'], unit='d')
Then, use pd.merge_asof, to merge your dataframe with the count of calls by each individual in a period of 7 days. To get this, use groupby with a pd.Grouper with a frequency of 7 days:
new_df = (pd.merge_asof(df.sort_values(['Day']),
df.sort_values(['Day'])
.groupby([pd.Grouper(key='Day', freq='7d'), 'PersonID'])
.size()
.to_frame('ContactsIN7Days')
.reset_index(),
left_on='Day', right_on='Day',
left_by='PersonID', right_by='PersonID',
direction='nearest'))
Your resulting new_df will look like this:
CallID Day PersonID ContactsIN7Days
0 5 1 days 3 2
1 6 2 days 3 2
2 1 8 days 1 1
3 2 12 days 2 2
4 3 14 days 2 2
5 7 100 days 3 1
I want to apply an operation on multiple groups of a data frame and then fill all values of that group by the result. Lets take mean and np.cumsum as an example and the following dataframe:
df=pd.DataFrame({"a":[1,3,2,4],"b":[1,1,2,2]})
which looks like this
a b
0 1 1
1 3 1
2 2 2
3 4 2
Now I want to group the dataframe by b, then take the mean of a in each group, then apply np.cumsum to the means, and then replace all values of a by the (group dependent) result.
For the first three steps, I would start like this
df.groupby("b").mean().apply(np.cumsum)
which gives
a
b
1 2
2 5
But what I want to get is
a b
0 2 1
1 2 1
2 5 2
3 5 2
Any ideas how this can be solved in a nice way?
You can use map by Series:
df1 = df.groupby("b").mean().cumsum()
print (df1)
a
b
1 2
2 5
df['a'] = df['b'].map(df1['a'])
print (df)
a b
0 2 1
1 2 1
2 5 2
3 5 2
I've got a "Perf" dataframe with people performance data over time.
The index is a timstamp and the columns are the persons name.
There are 100 persons (columns) and each person belongs to one of 10 groups, however the group assignment is dynamic, everyday each person could be assigned to a different group.
So there is a second "Group" DataFrame of the same shape than "Perf" that contains group number (0-9) for each timestamp and person.
The question is how can I elegantly do a mean subtraction everyday for each person with regards to its group assignment?
One method that is really slow is:
for g in range(10):
Perf[Group==g] -= Perf[Group==g].mean(1)
But this is really slow, I'm sure there is a way to do it in one shot with Pandas.
here is an concrete example:
scores represents the score for each person (0-4) for 10 days (0-9):
In [8]: perf = DataFrame(np.random.randn(10,5))
In [9]: perf
Out[9]:
0 1 2 3 4
0 0.945575 -0.805883 1.338865 0.420829 -1.074329
1 -1.086116 0.430230 1.296153 0.527612 1.269646
2 0.705276 -1.409828 2.859838 -0.769508 1.520295
3 0.331860 -0.217884 0.962576 -0.495888 -1.083996
4 0.402625 0.018885 -0.260516 -0.547802 -0.995959
5 2.168944 -0.361657 0.184537 0.391014 0.972161
6 1.959699 0.590739 -0.781736 1.059761 1.080997
7 2.090273 -2.446399 0.553785 0.806368 -0.786343
8 0.441160 -2.320302 -1.981387 2.190607 0.345626
9 -0.276013 -1.319214 1.339096 0.269680 -0.509884
Then I've got some grouping dataframe that for each day shows the group association of each one of the 5 persons, the grouping changes everyday.
In [20]: grouping
Out[20]:
0 1 2 3 4
0 3 1 2 1 2
1 3 1 2 2 1
2 2 2 3 1 1
3 1 2 2 3 1
4 3 2 1 2 1
5 2 1 1 2 3
6 1 2 1 2 3
7 2 2 1 1 3
8 2 1 2 1 3
9 1 3 2 1 2
I want to modify Perf such that for each day I subtract for each person the mean score of its group.
for example for day 0 it will be 0.0 -0.613356 1.206597 0.613356 -1.206597
I want to do it in 1 line without loops. Groupby seems to be the function to use, but I couldn't use efficiently its output to perform the mean subtract operation on the original matrix.