Reputation: 107
Lets us say there is a dataframe df
Name Balance
A 1000
B 5000
C 3000
D 6000
E 2000
F 5000
I am looking for an approach through which I can get three rows with highest balances among all.
df['balance'].get_indices_max(n=3) # where is no. of results required
Output when these indices will be used to get rows:
D 6000
F 5000
B 5000
UPDATE : EXTRA NOTES REGARDING THE ACCEPTED ANSWER
Possible "keep" values -
first : prioritize the first occurrence(s)
last : prioritize the last occurrence(s)
all : do not drop any duplicates, even it means selecting more than n items.
Upvotes: 0
Views: 87
Reputation: 1232
df = Df({"Name":list("ABCDEF"), "Balance":[1000,5000,3000,6000,2000,5000]})
index = df["Balance"].nlargest(3).index
df.loc[index]
Name Balance
3 D 6000
1 B 5000
5 F 5000
The columns that are not specified are returned as well, but not used for ordering. This method is equivalent to df.sort_values(columns, ascending=False).head(n), but more performant.
nlargest(3, keep='all')
keep{‘first’, ‘last’, ‘all’}, default ‘first’
When using keep='all', all duplicate items are maintained
Example
df = Df({"Name":list("ABCDEFX"), "Balance":[1000,5000,3000,6000,2000,5000,5000]})
index = df["Balance"].nlargest(3, keep='all').index
df.loc[index]
Name Balance
3 D 6000
1 B 5000
5 F 5000
6 X 5000
Upvotes: 3
Reputation: 323276
I usual do
out = df.sort_values('Balance').iloc[3:]
Out[476]:
Name Balance
1 B 5000
5 F 5000
3 D 6000
Upvotes: 1