I have a dataframe like in this one:
df = pd.DataFrame({'a':[1,2,1],'b':[4,6,0],'c':[0,4,8]})
+---+---+---+
| a | b | c |
+---+---+---+
| 1 | 4 | 0 |
+---+---+---+
| 2 | 6 | 4 |
+---+---+---+
| 1 | 0 | 8 |
+---+---+---+
for each row, I need (both) the 'n' (in this case two) highest values and the corresponding column in descending order:
row 1: 'b':4,'a':1
row 2: 'b':6,'c':4
row 3: 'c':8,'a':1
Here are two ways, both adapt from @unutbu's answer to "Find names of top-n highest-value columns in each pandas dataframe row"
1) Use Python Decorate-Sort-Undecorate with a .apply(lambda ...) on each row to insert the column names, do the np.argsort, keep the top-n, reformat the answer. (I think this is cleaner).
import numpy as np
# Apply Decorate-Sort row-wise to our df, and slice the top-n columns within each row...
sort_decr2_topn = lambda row, nlargest=2:
sorted(pd.Series(zip(df.columns, row)), key=lambda cv: -cv[1]) [:nlargest]
tmp = df.apply(sort_decr2_topn, axis=1)
0 [(b, 4), (a, 1)]
1 [(b, 6), (c, 4)]
2 [(c, 8), (a, 1)]
# then your result (as a pandas DataFrame) is...
np.array(tmp)
array([[('b', 4), ('a', 1)],
[('b', 6), ('c', 4)],
[('c', 8), ('a', 1)]], dtype=object)
# ... or as a list of rows is
tmp.values.tolist()
#... and you can insert the row-indices 0,1,2 with
zip(tmp.index, tmp.values.tolist())
[(0, [('b', 4), ('a', 1), ('c', 0)]), (1, [('b', 6), ('c', 4), ('a', 2)]), (2, [('c', 8), ('a', 1), ('b', 0)])]
2) Get the matrix of topnlocs as follows, then use it both to reindex into df.columns, and df.values, and combine that output:
import numpy as np
nlargest = 2
topnlocs = np.argsort(-df.values, axis=1)[:, 0:nlargest]
# ... now you can use topnlocs to reindex both into df.columns, and df.values, then reformat/combine them somehow
# however it's painful trying to apply that NumPy array of indices back to df or df.values,