Get Column and Row Index for Highest Value in Dataframe Pandas

Question

I'd like to know if there's a way to find the location (column and row index) of the highest value in a dataframe. So if for example my dataframe looks like this:

   A         B         C         D         E
0  100       9         1         12        6
1  80        10        67        15        91
2  20        67        1         56        23
3  12        51        5         10        58
4  73        28        72        25        1

How do I get a result that looks like this: [0, 'A'] using Pandas?

Answer 1

simple, fast, one liner:

loc = [df.max(axis=1).idxmax(), df.max().idxmax()]

(For large data frames, .stack() can be quite slow.)

Answer 2

print('Max value:', df.stack().max())
print('Parameters :', df.stack().idxmax())

This is the best way imho.

Answer 3

Use np.argmax

NumPy's argmaxcan be helpful:

>>> df.stack().index[np.argmax(df.values)]
(0, 'A')

In steps

df.values is a two-dimensional NumPy array:

>>> df.values
array([[100,   9,   1,  12,   6],
       [ 80,  10,  67,  15,  91],
       [ 20,  67,   1,  56,  23],
       [ 12,  51,   5,  10,  58],
       [ 73,  28,  72,  25,   1]])

argmax gives you the index for the maximum value for the "flattened" array:

>>> np.argmax(df.values)
0

Now, you can use this index to find the row-column location on the stacked dataframe:

>>> df.stack().index[0]
(0, 'A')

Fast Alternative

If you need it fast, do as few steps as possible. Working only on the NumPy array to find the indices np.argmax seems best:

v = df.values
i, j = [x[0] for x in np.unravel_index([np.argmax(v)], v.shape)]
[df.index[i], df.columns[j]]

Result:

[0, 'A']

Timings

Timing works best for lareg data frames:

df = pd.DataFrame(data=np.arange(int(1e6)).reshape(-1,5), columns=list('ABCDE'))

Sorted slowest to fastest:

Mask:

%timeit df.mask(~(df==df.max().max())).stack().index.tolist()
33.4 ms ± 982 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Stack-idmax

%timeit list(df.stack().idxmax())
17.1 ms ± 139 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Stack-argmax

%timeit df.stack().index[np.argmax(df.values)]
14.8 ms ± 392 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Where

%%timeit
i,j = np.where(df.values == df.values.max())
list((df.index[i].values.tolist()[0],df.columns[j].values.tolist()[0]))

4.45 ms ± 84.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Argmax-unravel_index

%%timeit

v = df.values
i, j = [x[0] for x in np.unravel_index([np.argmax(v)], v.shape)]
[df.index[i], df.columns[j]]

499 µs ± 12 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Compare

d = {'name': ['Mask', 'Stack-idmax', 'Stack-argmax', 'Where', 'Argmax-unravel_index'],
     'time': [33.4, 17.1, 14.8, 4.45, 499],
     'unit': ['ms', 'ms', 'ms', 'ms', 'µs']}


timings = pd.DataFrame(d)
timings['seconds'] = timings.time * timings.unit.map({'ms': 1e-3, 'µs': 1e-6})
timings['factor slower'] = timings.seconds / timings.seconds.min()
timings.sort_values('factor slower')

Output:

                   name    time unit   seconds  factor slower
4  Argmax-unravel_index  499.00   µs  0.000499       1.000000
3                 Where    4.45   ms  0.004450       8.917836
2          Stack-argmax   14.80   ms  0.014800      29.659319
1           Stack-idmax   17.10   ms  0.017100      34.268537
0                  Mask   33.40   ms  0.033400      66.933868

So the "Argmax-unravel_index" version seems to be one to nearly two orders of magnitude faster for large data frames, i.e. where often speeds matters most.

Answer 4

In my opinion for larger datasets, stack() becomes inefficient, let's use np.where to return index positions:

i,j = np.where(df.values == df.values.max())
list((df.index[i].values.tolist()[0],df.columns[j].values.tolist()[0]))

Output:

[0, 'A']

Timings for larger datafames:

df = pd.DataFrame(data=np.arange(10000).reshape(-1,5), columns=list('ABCDE'))

np.where method

> %%timeit i,j = np.where(df.values == df.values.max())
> list((df.index[i].values.tolist()[0],df.columns[j].values.tolist()[0]))

1000 loops, best of 3: 364 µs per loop

Other stack methods

> %timeit df.mask(~(df==df.max().max())).stack().index.tolist()

100 loops, best of 3: 7.68 ms per loop

> %timeit df.stack().index[np.argmax(df.values)`]

10 loops, best of 3: 50.5 ms per loop

> %timeit list(df.stack().idxmax())

1000 loops, best of 3: 1.58 ms per loop

Even larger dataframe:

df = pd.DataFrame(data=np.arange(100000).reshape(-1,5), columns=list('ABCDE'))

Respectively:

1000 loops, best of 3: 1.62 ms per loop
10 loops, best of 3: 18.2 ms per loop
100 loops, best of 3: 5.69 ms per loop
100 loops, best of 3: 6.64 ms per loop

Answer 5

Use stack for Series with MultiIndex and idxmax for index of max value:

print (df.stack().idxmax())
(0, 'A')

print (list(df.stack().idxmax()))
[0, 'A']

Detail:

print (df.stack())
0  A    100
   B      9
   C      1
   D     12
   E      6
1  A     80
   B     10
   C     67
   D     15
   E     91
2  A     20
   B     67
   C      1
   D     56
   E     23
3  A     12
   B     51
   C      5
   D     10
   E     58
4  A     73
   B     28
   C     72
   D     25
   E      1
dtype: int64

Answer 6

mask + max

df.mask(~(df==df.max().max())).stack().index.tolist()
Out[17]: [(0, 'A')]

Answer 7

This should work:

def max_df(df):
    m = None
    p = None
    for idx, item in enumerate(df.idxmax()):
        c = df.columns[item]
        val = df[c][idx]
        if m is None or val > m:
            m = val
            p = idx, c
    return p

This uses the idxmax function, then compares all of the values returned by it.

Example usage:

>>> df

     A  B
0  100  9
1   90  8
>>> max_df(df)

(0, 'A')

Here's a one-liner (for fun):

def max_df2(df):
    return max((df[df.columns[item]][idx], idx, df.columns[item]) for idx, item in enumerate(df.idxmax()))[1:]