Can I perform dynamic cumsum of rows in pandas?

问题

If I have the following dataframe, derived like so: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 1)))

Is there an efficient way cumsum rows with a limit and each time this limit is reached, to start a new cumsum. After each limit is reached (however many rows), a row is created with the total cumsum.

Below I have created an example of a function that does this, but it's very slow, especially when the dataframe becomes very large. I don't like that my function is looping and I am looking for a way to make it faster (I guess a way without a loop).

def foo(df, max_value):
    last_value = 0
    storage = []
    for index, row in df.iterrows():
        this_value = np.nansum([row[0], last_value])
        if this_value >= max_value:
            storage.append((index, this_value))
            this_value = 0
        last_value = this_value
    return storage

If you rum my function like so: foo(df, 5) In in the above context, it returns:

   0
2  10
6  8

回答1:

The loop cannot be avoided, but it can be parallelized using numba's njit:

from numba import njit, prange

@njit
def dynamic_cumsum(seq, index, max_value):
    cumsum = []
    running = 0
    for i in prange(len(seq)):
        if running > max_value:
            cumsum.append([index[i], running])
            running = 0
        running += seq[i] 
    cumsum.append([index[-1], running])

    return cumsum

The index is required here, assuming your index is not numeric/monotonically increasing.

%timeit foo(df, 5)
1.24 ms ± 41.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit dynamic_cumsum(df.iloc(axis=1)[0].values, df.index.values, 5)
77.2 µs ± 4.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

If the index is of Int64Index type, you can shorten this to:

@njit
def dynamic_cumsum2(seq, max_value):
    cumsum = []
    running = 0
    for i in prange(len(seq)):
        if running > max_value:
            cumsum.append([i, running])
            running = 0
        running += seq[i] 
    cumsum.append([i, running])

    return cumsum

lst = dynamic_cumsum2(df.iloc(axis=1)[0].values, 5)
pd.DataFrame(lst, columns=['A', 'B']).set_index('A')

    B
A    
3  10
7   8
9   4

%timeit foo(df, 5)
1.23 ms ± 30.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit dynamic_cumsum2(df.iloc(axis=1)[0].values, 5)
71.4 µs ± 1.4 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

njit Functions Performance

perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.randint(0, 10, size=(n, 1))),
    kernels=[
        lambda df: list(cumsum_limit_nb(df.iloc[:, 0].values, 5)),
        lambda df: dynamic_cumsum2(df.iloc[:, 0].values, 5)
    ],
    labels=['cumsum_limit_nb', 'dynamic_cumsum2'],
    n_range=[2**k for k in range(0, 17)],
    xlabel='N',
    logx=True,
    logy=True,
    equality_check=None # TODO - update when @jpp adds in the final `yield`
)

The log-log plot shows that the generator function is faster for larger inputs:

A possible explanation is that, as N increases, the overhead of appending to a growing list in dynamic_cumsum2 becomes prominent. While cumsum_limit_nb just has to yield.

回答2:

A loop isn't necessarily bad. The trick is to make sure it's performed on low-level objects. In this case, you can use Numba or Cython. For example, using a generator with numba.njit:

from numba import njit

@njit
def cumsum_limit(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

idx, vals = zip(*cumsum_limit(df[0].values))
res = pd.Series(vals, index=idx)

To demonstrate the performance benefits of JIT-compiling with Numba:

import pandas as pd, numpy as np
from numba import njit

df = pd.DataFrame({0: [0, 2, 8, 1, 0, 0, 7, 0, 2, 2]})

@njit
def cumsum_limit_nb(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

def cumsum_limit(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

n = 10**4
df = pd.concat([df]*n, ignore_index=True)

%timeit list(cumsum_limit_nb(df[0].values))  # 4.19 ms ± 90.4 µs per loop
%timeit list(cumsum_limit(df[0].values))     # 58.3 ms ± 194 µs per loop

回答3:

simpler approach:

def dynamic_cumsum(seq,limit):
    res=[]
    cs=seq.cumsum()
    for i, e in enumerate(cs):
        if cs[i] >limit:
            res.append([i,e])
            cs[i+1:] -= e
    if res[-1][0]==i:
        return res
    res.append([i,e])
    return res

result:

x=dynamic_cumsum(df[0].values,5)
x
>>[[2, 10], [6, 8], [9, 4]]

来源：https://stackoverflow.com/questions/54208023/can-i-perform-dynamic-cumsum-of-rows-in-pandas

标签

python

pandas

performance

numba