CODEWITHSUNDEEP
pandas
NinjaGaiden
NinjaGaiden

Reputation: 3146

select column index by percentage of last row

I am trying to extract the top % of the last row value in dataframe. 

df=pd.DataFrame.from_dict(
{
'u':[6,7,11],
'v':[4,7,1],
'w':[0,4,7],
'x':[1,3,5],
'y':[2,1,6],
'z':[3,9,8],
})

pct=40 #40%
val=int((pct*.01)*len(df.columns))
s=df.iloc[-1].nlargest(val)

print (df[s.keys())

I was wondering if this is the correct way.

Upvotes: 0

Views: 68

Answers (1)

jezrael
jezrael

Reputation: 863741

I believe you need:

pct=40 #40%
val=int((pct*.01)*len(df.columns))
df = df.loc[:, df.iloc[-1].nlargest(val).index]
print (df)
    u  z
0   6  3
1   7  9
2  11  8

Solution with numpy.argsort for positions by sorted data, select last columns in inverse order and pass to iloc:

pct=40 #40%
val=int((pct*.01)*len(df.columns))

df = df.iloc[:, df.iloc[-1].argsort()[:-val - 1:-1]]
print (df)
    u  z
0   6  3
1   7  9
2  11  8

Upvotes: 1

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