Simple, Fast, and Pandaic: ngroups Newer versions of the groupby API (pandas >= 0.23) provide this (undocumented) attribute which stores the number of groups in a GroupBy object. # setup df = pd.DataFrame({‘A’: list(‘aabbcccd’)}) dfg = df.groupby(‘A’) # call `.ngroups` on the GroupBy object dfg.ngroups # 4 Note that this is different from GroupBy.groups which … Read more
You should be able to achieve what you want using the groupby / apply pattern. The below code should be helpful. Create example data: from statsmodels.regression.rolling import RollingOLS from statsmodels.tools.tools import add_constant import pandas as pd import numpy as np # make some toy data race_dates = pd.to_datetime([‘2020-06-09’]*3 + [‘2020-12-01’]*4 + [‘2021-01-21’]*4 + [‘2021-05-04’]*5) distance … Read more
Copy Data from OP and run df = pd.read_clipboard() Plot using pandas.DataFrame.plot Updated to pandas v1.2.4 and matplotlib v3.3.4 then using your code df = df.replace(np.nan, 0) dfg = df.groupby([‘home_team’])[‘arrests’].mean() dfg.plot(kind=’bar’, title=”Arrests”, ylabel=”Mean Arrests”, xlabel=”Home Team”, figsize=(6, 5))
Use groupby + agg by dict, so then is necessary order columns by subset or reindex_axis. Last add reset_index for convert index to column if necessary. df = a.groupby(‘user’).agg({‘num1′:’min’, ‘num2′:’max’})[[‘num1′,’num2’]].reset_index() print (df) user num1 num2 0 a 1 3 1 b 4 5 What is same as: df = a.groupby(‘user’).agg({‘num1′:’min’, ‘num2′:’max’}) .reindex_axis([‘num1′,’num2’], axis=1) .reset_index() print … Read more
By using apply df.groupby([‘col1’, ‘col2’])[“col3”, “col4”].apply(lambda x : x.astype(int).sum()) Out[1257]: col3 col4 col1 col2 a c 2 4 d 1 2 b d 1 2 e 2 4 If you want to agg df.groupby([‘col1’, ‘col2’]).agg({‘col3′:’sum’,’col4′:’sum’})
apply applies the function to each group (your Species). Your function returns 1, so you end up with 1 value for each of 3 groups. agg aggregates each column (feature) for each group, so you end up with one value per column per group. Do read the groupby docs, they’re quite helpful. There are also … Read more
consider the dataframe df df = pd.DataFrame(dict(A=list(‘aabb’), B=[1, 2, 3, 4], C=[0, 9, 0, 9])) groupby is the standard use aggregater df.groupby(‘A’).mean() maybe you want these values broadcast across the whole group and return something with the same index as what you started with. use transform df.groupby(‘A’).transform(‘mean’) df.set_index(‘A’).groupby(level=”A”).transform(‘mean’) agg is used when you have specific … Read more
First groupby the key1 column: In [11]: g = df.groupby(‘key1′) and then for each group take the subDataFrame where key2 equals ‘one’ and sum the data1 column: In [12]: g.apply(lambda x: x[x[‘key2’] == ‘one’][‘data1′].sum()) Out[12]: key1 a 0.093391 b 1.468194 dtype: float64 To explain what’s going on let’s look at the ‘a’ group: In [21]: … Read more
Use value_counts with normalize=True: df[‘gender’].value_counts(normalize=True) * 100 The result is a fraction in range (0, 1]. We multiply by 100 here in order to get the %.
Apply a lambda and call sample with param frac: In [2]: df = pd.DataFrame({‘a’: [1,2,3,4,5,6,7], ‘b’: [1,1,1,0,0,0,0]}) grouped = df.groupby(‘b’) grouped.apply(lambda x: x.sample(frac=0.3)) Out[2]: a b b 0 6 7 0 1 2 3 1