Reputation: 2080
I have a DataFrame I wanted the difference between the maximum and second maximum from the DataFrame as a new column appended to the DataFrame as output.
The data frame looks like this for example (this is quite a huge DataFrame):
gene_id Time_1 Time_2 Time_3
a 0.01489251 8.00246 8.164309
b 6.67943235 0.8832114 1.048761
So far I tried the following but it's just taking the headers,
largest = max(df)
second_largest = max(item for item in df if item < largest)
and returning the header value alone.
Upvotes: 3
Views: 1428
Reputation: 8906
Here's an elegant solution that doesn't involve sorting or defining any functions. It's also fully vectorized as it avoid use of the apply
method.
maxes = df.max(axis=1)
less_than_max = df.where(df.lt(maxes, axis='rows'))
seconds = less_than_max.max(axis=1)
df['diff'] = maxes - seconds
Upvotes: 1
Reputation: 394031
You can define a func which takes the values, sorts them, slices the top 2 values ([:2]
) then calculates the difference and returns the second value (as the first value is NaN
). You apply
this and pass arg axis=1
to apply row-wise:
In [195]:
def func(x):
return -x.sort(inplace=False, ascending=False)[:2].diff()[1]
df['diff'] = df.loc[:,'Time_1':].apply(func, axis=1)
df
Out[195]:
gene_id Time_1 Time_2 Time_3 diff
0 a 0.014893 8.002460 8.164309 0.161849
1 b 6.679432 0.883211 1.048761 5.630671
Upvotes: 1
Reputation: 9798
Here is my solution:
# Load data
data = {'a': [0.01489251, 8.00246, 8.164309], 'b': [6.67943235, 0.8832114, 1.048761]}
df = pd.DataFrame.from_dict(data, 'index')
The trick is to do a linear sort of the values and keep the top-2 using numpy.argpartition. You do the difference of the 2 maximum values in absolute value. The function is applied row-wise.
def f(x):
ind = np.argpartition(x.values, -2)[-2:]
return np.abs(x.iloc[ind[0]] - x.iloc[ind[1]])
df.apply(f, axis=1)
Upvotes: 1