Issue
I'm a beginner with Pandas. I've got two dataframes df1 and df2 of three columns each, labelled by some index.
I would like to get a third dataframe whose entries are min( df1-df2, 1-df1-df2 ) for each column, while preserving the index.
I don't know how to do this on all the three columns at once. If I try e.g. np.min( df1-df2, 1-df1-df2 )
I get TypeError: 'DataFrame' objects are mutable, thus they cannot be hashed
, whereas min( df1-df2, 1-df1+df2 )
gives ValueError: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I can't use apply
because I've got more than one dataframe. Basically, I would like to use something like subtract
, but with the ability to define my own function.
Example: consider these two dataframes
df0 = pd.DataFrame( [[0.1,0.2,0.3], [0.3, 0.1, 0.2], [0.1, 0.3, 0.9]], index=[2,1,3], columns=['px', 'py', 'pz'] )
In [4]: df0
Out[4]:
px py pz
2 0.1 0.2 0.3
1 0.3 0.1 0.2
3 0.1 0.3 0.9
and
df1 = pd.DataFrame( [[0.9,0.1,0.9], [0.1,0.2,0.1], [0.3,0.1,0.8]], index=[3,1,2], columns=['px', 'py', 'pz'])
px py pz
3 0.9 0.1 0.9
1 0.1 0.2 0.1
2 0.3 0.1 0.8
my desired output is a new dataframe df, made up of three columns 'px', 'py', 'pz', whose entries are:
for j in range(1,4):
dfx[j-1] = min( df0['px'][j] - df1['px'][j], 1 - df0['px'][j] + df1['px'][j] )
for df['px'], and similarly for 'py' and 'pz'.
px py pz
1 0.2 -0.1 0.1
2 -0.2 0.1 -0.5
3 -0.8 0.2 0.0
I hope it's clear now! Thanks in advance!
Solution
pandas
is smart enough to match up the columns and index values for you in a vectorized way. If you're looping a dataframe, you're probably doing it wrong.
m1 = df0 - df1
m2 = 1 - (df0 + df1)
# Take the values from m1 where they're less than
# The corresponding value in m2. Otherwise, take m2:
out = m1[m1.lt(m2)].combine_first(m2)
# Another method: Combine our two calculated frames,
# groupby the index, and take the minimum.
out = pd.concat([m1, m2]).groupby(level=0).min()
print(out)
# Output:
px py pz
1 0.2 -0.1 0.1
2 -0.2 0.1 -0.5
3 -0.8 0.2 -0.8
Answered By - BeRT2me
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