Efficient pandas/numpy function for time since change

问题

Given a Series , I would like to efficiently compute how many observations have passed since there was a change. Here is a simple example:

ser = pd.Series([1.2,1.2,1.2,1.2,2,2,2,4,3])

print(ser)

0    1.2
1    1.2
2    1.2
3    1.2
4    2.0
5    2.0
6    2.0
7    4.0
8    3.0

I would like to apply a function to ser which would result in:

As I am dealing with large series I would prefer a fast solution that does not involve looping. Thanks

Edit If possible, would like the function to work also for series with identical values (which would just result in a series of integers incremented by 1)

回答1:

Here's one NumPy approach -

def array_cumcount(a):
    idx = np.flatnonzero(a[1:] != a[:-1])+1
    shift_arr = np.ones(a.size,dtype=int)
    shift_arr[0] = 0

    if len(idx)>=1:
        shift_arr[idx[0]] = -idx[0]+1
        shift_arr[idx[1:]] = -idx[1:] + idx[:-1] + 1
    return shift_arr.cumsum()

Sample run -

In [583]: ser = pd.Series([1.2,1.2,1.2,1.2,2,2,2,4,3,3,3,3])

In [584]: array_cumcount(ser.values)
Out[584]: array([0, 1, 2, 3, 0, 1, 2, 0, 0, 1, 2, 3])

Runtime test -

In [601]: ser = pd.Series(np.random.randint(0,3,(10000)))

# @Psidom's soln
In [602]: %timeit ser.groupby(ser).cumcount()
1000 loops, best of 3: 729 µs per loop

In [603]: %timeit array_cumcount(ser.values)
10000 loops, best of 3: 85.3 µs per loop

In [604]: ser = pd.Series(np.random.randint(0,3,(1000000)))

# @Psidom's soln
In [605]: %timeit ser.groupby(ser).cumcount()
10 loops, best of 3: 30.1 ms per loop

In [606]: %timeit array_cumcount(ser.values)
100 loops, best of 3: 11.7 ms per loop

回答2:

You can use groupby.cumcount:

ser.groupby(ser).cumcount()

#0    0
#1    1
#2    2
#3    3
#4    0
#5    1
#6    2
#7    0
#8    0
#dtype: int64

来源：https://stackoverflow.com/questions/43211261/efficient-pandas-numpy-function-for-time-since-change

标签

python-3.x

pandas

numpy

series