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I need to fill a column with values, that are present in a set and not present in any other columns.

initial df

 c0 c1 c2 c3 c4 c5 0 4 5 6 3 2 1 1 1 5 4 0 2 3 2 5 6 4 0 1 3 3 5 4 6 2 0 1 4 5 6 4 0 1 3 5 0 1 4 5 6 2

I need df['c6'] column that is a set-like difference operation product between a set of set([0,1,2,3,4,5,6]) and each row of df

so that the result df is

 c0 c1 c2 c3 c4 c5 c6 0 4 5 6 3 2 1 0 1 1 5 4 0 2 3 6 2 5 6 4 0 1 3 2 3 5 4 6 2 0 1 3 4 5 6 4 0 1 3 2 5 0 1 4 5 6 2 3

Thank you!

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3 Answers 3

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2

Slightly different approach:

df['c6'] = sum(range(7)) - df.sum(axis=1)

or if you want to be more verbose:

df['c6'] = sum([0,1,2,3,4,5,6]) - df.sum(axis=1)
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2 Comments

but would scale quite well for large dataframes because of vectorisation, I think.
though this approach is not universal, it works best for my particular problem. Thank you!
0

Use numpy setdiff1d to find the difference between the two arrays and assign the output to column c6 

ck = np.array([0,1,2,3,4,5,6]) M = df.to_numpy() df['c6'] = [np.setdiff1d(ck,i)[0] for i in M] c0 c1 c2 c3 c4 c5 c6 0 4 5 6 3 2 1 0 1 1 5 4 0 2 3 6 2 5 6 4 0 1 3 2 3 5 4 6 2 0 1 3 4 5 6 4 0 1 3 2 5 0 1 4 5 6 2 3
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A simple way I could think of is using a list comprehension and set difference:

s = {0, 1, 2, 3, 4, 5, 6} s {0, 1, 2, 3, 4, 5, 6} df['c6'] = [tuple(s.difference(vals))[0] for vals in df.values] df c0 c1 c2 c3 c4 c5 c6 0 4 5 6 3 2 1 0 1 1 5 4 0 2 3 6 2 5 6 4 0 1 3 2 3 5 4 6 2 0 1 3 4 5 6 4 0 1 3 2 5 0 1 4 5 6 2 3
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