按列总和对熊猫数据框进行排序

Question

I have a dataframe that looks like this

            Australia  Austria    United Kingdom  Vietnam
date                                                    
2020-01-30          9        0                 1       2
2020-01-31          9        9                 4       2

I would like to crate a new dataframe that inclues countries that have sum of their column > 4 and I do it

df1 = df[[i for i in df.columns if int(df[i].sum()) > 4]]

this gives me

            Australia  Austria    United Kingdom  
date                                                     
2020-01-30          9        0                 1      
2020-01-31          9        9                 4 

I now would like to sort the countries based on the sum of their column and than take the first 2

            Australia  Austria   
date                                    
2020-01-30          9        0        
2020-01-31          9        9

I know I have to use sort_values and tail . I just can't workout how

Answer 1

IIUC, you can do:

s = df.sum()
df[s.sort_values(ascending=False).index[:2]]

Output:

            Australia  Austria
date                          
2020-01-30          9        0
2020-01-31          9        9

Answer 2

First filter for sum greater like 4 and then add Series.nlargest for top2 sum and filter by index values:

s = df.sum()

df = df[s[s > 4].nlargest(2).index]
print (df)
            Australia  Austria
date                          
2020-01-30          9        0
2020-01-31          9        9

Details :

print (s)
Australia    18.0
Austria       9.0
United        5.0
Kingdom       4.0
Vietnam       0.0
dtype: float64

print (s[s > 4])
Australia    18.0
Austria       9.0
United        5.0
dtype: float64

print (s[s > 4].nlargest(2))
Australia    18.0
Austria       9.0
dtype: float64

print (s[s > 4].nlargest(2).index)
Index(['Australia', 'Austria'], dtype='object')

Answer 3

You can take the sum of the dataframe along the first axis, sort_values and take the first n columns:

df[df.sum(0).sort_values(ascending=False)[:2].index]


               Australia  Austria
2020-01-30          9        0
2020-01-31          9        9

Answer 4

another way modifying your list comp slightly.

cols = df[[i for i in df.columns if int(df[i].sum()) > 4]].stack().groupby(level=1).sum().head(2).index

#would yield the same result df.stack().groupby(level=1).sum().head(2).index


df[cols]

            Australia  Austria
date                          
2020-01-30          9        0
2020-01-31          9        9

Answer 5

You can also do this inline using the .pipe function, which helps if you don't want to define a variable for a temporary result:

df.pipe(lambda df: df.loc[:, df.sum().sort_values(ascending=False).index])

For example, you might have a pipeline:

new_df = (
    df1
    # Some example operations one might do:
    .groupby('column')
    .apply(sum).unstack()
    .fillna(0).astype(int)
    # Sort columns by total count:
    .pipe(lambda df: df.loc[:, df.sum().sort_values(ascending=False).index])
)