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Pandas get count of group above group median


Pandas get count of group above group median

By : shashikala rs
Date : October 18 2020, 01:08 AM
may help you . Given a Pandas dataframe with categorical column family and numerical column score, I want to get the count of those by family who have above (or below) the median score for that family. , It sounds like you are looking for something like
code :
df[df.score > df.groupby('family').score.transform('median')].groupby('family').count()


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Python - Pandas - Plotting Count of Column by Group - Graphing Each Group Over Time

Python - Pandas - Plotting Count of Column by Group - Graphing Each Group Over Time


By : user2215976
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Sorry if the title is horribly vague, its hard to express the issue in a few words.
code :
d = {'level' : ['ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR'], 
 'DATE' : ['2014-07-29 12:35:55.916', '2014-07-29 12:35:55.916', '2014-07-29 12:35:55.916', '2014-07-29 12:35:55.874', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.975'],
 'APP' : ['app1', 'app1', 'app1', 'app2', 'app3', 'app3', 'app3', 'app3', 'app3', 'app4']}
df = pd.DataFrame(d)
df.pivot_table(values='level', index='DATE', columns='APP', aggfunc=len).plot(kind='bar')
How to group by and count number of none missing values for each column in group in pandas

How to group by and count number of none missing values for each column in group in pandas


By : RX178 MK2
Date : March 29 2020, 07:55 AM
like below fixes the issue You can use DF.GroupBy.count which includes only Non-NaN entries while counting. So, you can let var be the grouped key and then aggregate the counts respectively for the two selected columns of the DF as shown:
code :
cols = ['qualified_date', 'loyal_date']
df.groupby('var')[cols].agg('count').add_suffix("_count").reset_index()
how to group by date and unique each group and count each group with pandas

how to group by date and unique each group and count each group with pandas


By : Fury
Date : March 29 2020, 07:55 AM
Hope this helps how to group by date and unique each group and count each group with pandas? , I believe need add ['Client MAC'].nunique():
code :
df = (pd.concat([df[['date','Client MAC']], 
           df8[['date',"MAC address"]].rename(columns={"MAC address":"Client MAC"})])
        .groupby(["date"])['Client MAC']
        .nunique())
df = (pd.concat([df[['date','Client MAC']], 
                df8[['date',"MAC address"]].rename(columns={"MAC address":"Client MAC"})]))

df = df['Client MAC'].groupby(df["date"].dt.date).nunique()
How to count number of rows per group greater than the average of that group in pandas group by?

How to count number of rows per group greater than the average of that group in pandas group by?


By : mrfazeli
Date : March 29 2020, 07:55 AM
Any of those help I have a data frame df and I use several columns from it to groupby: , You could also aggregate using a lambda function as the following:
code :
df.groupby(['col1', 'col2']).agg(['mean', 'count', 
                                  lambda x: (x > x.mean()).sum()])
Pandas: How to group by one column and show count for unique values for all other columns per group?

Pandas: How to group by one column and show count for unique values for all other columns per group?


By : soren
Date : March 29 2020, 07:55 AM
Hope this helps Use DataFrame.melt with DataFrame.pivot_table and flatten MultiIndex:
code :
df = df.melt('Decision').pivot_table(index='Decision', 
                                     columns=['variable','value'], 
                                     aggfunc='size', 
                                     fill_value=0)
df.columns = df.columns.map('{0[0]}_{0[1]}'.format)
df = df.reset_index()
print (df)
  Decision  Area_North  Area_South  Class_1  Class_2  Class_3  Type_A  Type_B  \
0       No           1           4        2        1        2       1       2   
1      Yes           3           1        2        1        1       2       2   

   Type_C  
0       2  
1       0  
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