Finding top N columns for each row in data frame

Question

given a data frame with one descriptive column and X numeric columns, for each row I\'d like to identify the top N columns with the higher values and save it as rows on a ne

Answer 1

Yet another crazy one-liner, given n = 3

{index:option for (index, option) in zip(df['index'], 
    [df.columns[pd.notnull(x[1].where(x[1][1:].sort_values()[-n:]))].tolist()
        for x in df.iterrows()])}

{'A': ['option2', 'option3', 'option4'],
 'C': ['option2', 'option4', 'option5'],
 'B': ['option1', 'option3', 'option4'],
 'E': ['option1', 'option2', 'option3'],
 'D': ['option1', 'option2', 'option5'],
 'F': ['option1', 'option3', 'option5']}

Answer 2

Let's assume

N = 3

First of all I will create matrix of input fields and for each field remember what was original option for this cell:

matrix = [[(j, 'option' + str(i)) for j in df['option' + str(i)]] for i in range(1,6)]

The result of this line will be:

[
 [(1, 'option1'), (5, 'option1'), (3, 'option1'), (7, 'option1'), (9, 'option1'), (3, 'option1')],
 [(8, 'option2'), (4, 'option2'), (5, 'option2'), (6, 'option2'), (9, 'option2'), (2, 'option2')],
 [(9, 'option3'), (9, 'option3'), (1, 'option3'), (3, 'option3'), (9, 'option3'), (5, 'option3')],
 [(3, 'option4'), (8, 'option4'), (3, 'option4'), (5, 'option4'), (7, 'option4'), (0, 'option4')],
 [(2, 'option5'), (3, 'option5'), (4, 'option5'), (9, 'option5'), (4, 'option5'), (2, 'option5')]
]

Then we can easly transform matrix using zip function, sort result rows by first element of tuple and take N first items:

transformed = [sorted(l, key=lambda x: x[0], reverse=True)[:N] for l in zip(*matrix)]

List transformed will look like:

[
 [(9, 'option3'), (8, 'option2'), (3, 'option4')],
 [(9, 'option3'), (8, 'option4'), (5, 'option1')],
 [(5, 'option2'), (4, 'option5'), (3, 'option1')],
 [(9, 'option5'), (7, 'option1'), (6, 'option2')],
 [(9, 'option1'), (9, 'option2'), (9, 'option3')],
 [(5, 'option3'), (3, 'option1'), (2, 'option2')]
]

The last step will be joining column index and result tuple by:

for id, top in zip(df['index'], transformed):
    for option in top:
        print id + ',' + option[1]
    print ''

Answer 3

This might not be so elegant, but I think it pretty much gets what you want:

n = 3
df.index = pd.Index(df['index'])
del df['index']
df = df.transpose().unstack()
for i, g in df.groupby(level=0):
    g = g.sort_values(ascending=False)
    print i, list(g.index.get_level_values(1)[:n])

Answer 4

If you just want pairings:

from operator import itemgetter as it
from itertools import repeat
n = 3

 # sort_values = order pandas < 0.17
new_d = (zip(repeat(row["index"]), map(it(0),(row[1:].sort_values(ascending=0)[:n].iteritems())))
                 for _, row in df.iterrows())
for row in new_d:
    print(list(row))

Output:

[('B', 'option3'), ('B', 'option4'), ('B', 'option1')]
[('C', 'option2'), ('C', 'option5'), ('C', 'option1')]
[('D', 'option5'), ('D', 'option1'), ('D', 'option2')]
[('E', 'option1'), ('E', 'option2'), ('E', 'option3')]
[('F', 'option3'), ('F', 'option1'), ('F', 'option2')]

Which also maintains the order.

If you want a list of lists:

from operator import itemgetter as it
from itertools import repeat
n = 3

new_d = [list(zip(repeat(row["index"]), map(it(0),(row[1:].sort_values(ascending=0)[:n].iteritems()))))
                 for _, row in df.iterrows()]

Output:

[[('A', 'option3'), ('A', 'option2'), ('A', 'option4')],
[('B', 'option3'), ('B', 'option4'), ('B', 'option1')], 
[('C', 'option2'), ('C', 'option5'), ('C', 'option1')], 
[('D', 'option5'), ('D', 'option1'), ('D', 'option2')], 
[('E', 'option1'), ('E', 'option2'), ('E', 'option3')],
[('F', 'option3'), ('F', 'option1'), ('F', 'option2')]]

Or using pythons sorted:

new_d = [list(zip(repeat(row["index"]), map(it(0), sorted(row[1:].iteritems(), key=it(1) ,reverse=1)[:n])))
                     for _, row in df.iterrows()]

Which is actually the fastest, if you really want strings, it is pretty trivial to format the output however you want.

Answer 5

dfc = df.copy()
result = {}

#First, I would effectively transpose this

for key in dfc:
    if key != 'index':
        for i in xrange(0,len(dfc['index'])):
            if dfc['index'][i] not in result:
                result[dfc['index'][i]] = []
            result[dfc['index'][i]] += [(key,dfc[key][i])]


def get_topn(result,n):
    #Use this to get the top value for each option
    return [x[0] for x in sorted(result,key=lambda x:-x[1])[0:min(len(result),n)]]


#Lastly, print the output in your desired format.
n = 3
keys = sorted([k for k in result])
for key in keys:
      for option in get_topn(result[key],n):
         print str(key) + ',' + str(option)
      print