Pandas: create two new columns in a dataframe with values calculated from a pre-existing column


Question

I am working with the pandas library and I want to add two new columns to a dataframe df with n columns (n > 0).
These new columns result from the application of a function to one of the columns in the dataframe.

The function to apply is like:

def calculate(x):
    ...operate...
    return z, y

One method for creating a new column for a function returning only a value is:

df['new_col']) = df['column_A'].map(a_function)

So, what I want, and tried unsuccesfully (*), is something like:

(df['new_col_zetas'], df['new_col_ys']) = df['column_A'].map(calculate)

What the best way to accomplish this could be ? I scanned the documentation with no clue.

**df['column_A'].map(calculate) returns a pandas Series each item consisting of a tuple z, y. And trying to assign this to two dataframe columns produces a ValueError.*

1
91
8/23/2018 2:11:40 PM

Accepted Answer

I'd just use zip:

In [1]: from pandas import *

In [2]: def calculate(x):
   ...:     return x*2, x*3
   ...: 

In [3]: df = DataFrame({'a': [1,2,3], 'b': [2,3,4]})

In [4]: df
Out[4]: 
   a  b
0  1  2
1  2  3
2  3  4

In [5]: df["A1"], df["A2"] = zip(*df["a"].map(calculate))

In [6]: df
Out[6]: 
   a  b  A1  A2
0  1  2   2   3
1  2  3   4   6
2  3  4   6   9
107
9/10/2012 5:20:49 PM

The top answer is flawed in my opinion. Hopefully, no one is mass importing all of pandas into their namespace with from pandas import *. Also, the map method should be reserved for those times when passing it a dictionary or Series. It can take a function but this is what apply is used for.

So, if you must use the above approach, I would write it like this

df["A1"], df["A2"] = zip(*df["a"].apply(calculate))

There's actually no reason to use zip here. You can simply do this:

df["A1"], df["A2"] = calculate(df['a'])

This second method is also much faster on larger DataFrames

df = pd.DataFrame({'a': [1,2,3] * 100000, 'b': [2,3,4] * 100000})

DataFrame created with 300,000 rows

%timeit df["A1"], df["A2"] = calculate(df['a'])
2.65 ms ± 92.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit df["A1"], df["A2"] = zip(*df["a"].apply(calculate))
159 ms ± 5.24 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

60x faster than zip


In general, avoid using apply

Apply is generally not much faster than iterating over a Python list. Let's test the performance of a for-loop to do the same thing as above

%%timeit
A1, A2 = [], []
for val in df['a']:
    A1.append(val**2)
    A2.append(val**3)

df['A1'] = A1
df['A2'] = A2

298 ms ± 7.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

So this is twice as slow which isn't a terrible performance regression, but if we cythonize the above, we get much better performance. Assuming, you are using ipython:

%load_ext cython

%%cython
cpdef power(vals):
    A1, A2 = [], []
    cdef double val
    for val in vals:
        A1.append(val**2)
        A2.append(val**3)

    return A1, A2

%timeit df['A1'], df['A2'] = power(df['a'])
72.7 ms ± 2.16 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Directly assigning without apply

You can get even greater speed improvements if you use the direct vectorized operations.

%timeit df['A1'], df['A2'] = df['a'] ** 2, df['a'] ** 3
5.13 ms ± 320 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

This takes advantage of NumPy's extremely fast vectorized operations instead of our loops. We now have a 30x speedup over the original.


The simplest speed test with apply

The above example should clearly show how slow apply can be, but just so its extra clear let's look at the most basic example. Let's square a Series of 10 million numbers with and without apply

s = pd.Series(np.random.rand(10000000))

%timeit s.apply(calc)
3.3 s ± 57.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Without apply is 50x faster

%timeit s ** 2
66 ms ± 2 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

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