加快python中的迭代器操作

Question

[pd.Series(pd.date_range(row[1].START_DATE, row[1].END_DATE)) for row in df[['START_DATE', 'END_DATE']].iterrows()]

無論如何，有沒有要加快這項作業的速度？ 基本上對於給定的日期范圍，我正在它們之間創建所有日期行。

Answer 1

使用DataFrame.itertuples ：

L = [pd.Series(pd.date_range(r.START_DATE, r.END_DATE)) for r in df.itertuples()]

或兩列的zip：

L = [pd.Series(pd.date_range(s, e)) for s, e in zip(df['START_DATE'], df['END_DATE'])]

如果要加入：

s = pd.concat(L, ignore_index=True)

100行的性能：

np.random.seed(123)

def random_dates(start, end, n=100):

    start_u = start.value//10**9
    end_u = end.value//10**9

    return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

start = pd.to_datetime('2015-01-01')
end = pd.to_datetime('2018-01-01')
df = pd.DataFrame({'START_DATE': start, 'END_DATE':random_dates(start, end)})
print (df)

---

In [155]: %timeit [pd.Series(pd.date_range(row[1].START_DATE, row[1].END_DATE)) for row in df[['START_DATE', 'END_DATE']].iterrows()]
33.5 ms ± 145 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [156]: %timeit [pd.date_range(row[1].START_DATE, row[1].END_DATE) for row in df[['START_DATE', 'END_DATE']].iterrows()]
30.3 ms ± 1.91 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [157]: %timeit [pd.Series(pd.date_range(r.START_DATE, r.END_DATE)) for r in df.itertuples()]
25.3 ms ± 218 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [158]: %timeit [pd.Series(pd.date_range(s, e)) for s, e in zip(df['START_DATE'], df['END_DATE'])]
24.3 ms ± 594 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

對於1000行：

start = pd.to_datetime('2015-01-01')
end = pd.to_datetime('2018-01-01')
df = pd.DataFrame({'START_DATE': start, 'END_DATE':random_dates(start, end, n=1000)})

In [159]: %timeit [pd.Series(pd.date_range(row[1].START_DATE, row[1].END_DATE)) for row in df[['START_DATE', 'END_DATE']].iterrows()]
333 ms ± 3.32 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [160]: %timeit [pd.date_range(row[1].START_DATE, row[1].END_DATE) for row in df[['START_DATE', 'END_DATE']].iterrows()]
314 ms ± 36.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [161]: %timeit [pd.Series(pd.date_range(s, e)) for s, e in zip(df['START_DATE'], df['END_DATE'])]
243 ms ± 1.49 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [162]: %timeit [pd.Series(pd.date_range(r.START_DATE, r.END_DATE)) for r in df.itertuples()]
246 ms ± 2.93 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 2

不用在每次迭代中創建pd.Series ，而是：

[pd.date_range(row[1].START_DATE, row[1].END_DATE))
 for row in df[['START_DATE', 'END_DATE']].iterrows()]

並根據結果創建一個數據框。 這是一個例子：

df = pd.DataFrame([
     {'start_date': pd.datetime(2019,1,1), 'end_date': pd.datetime(2019,1,10)},
     {'start_date': pd.datetime(2019,1,2), 'end_date': pd.datetime(2019,1,8)}, 
     {'start_date': pd.datetime(2019,1,6), 'end_date': pd.datetime(2019,1,14)} 
])

dr = [pd.date_range(df.loc[i,'start_date'], df.loc[i,'end_date']) for i,_ in df.iterrows()]

pd.DataFrame(dr)

      0          1          2          3          4          5  \
0 2019-01-01 2019-01-02 2019-01-03 2019-01-04 2019-01-05 2019-01-06   
1 2019-01-02 2019-01-03 2019-01-04 2019-01-05 2019-01-06 2019-01-07   
2 2019-01-06 2019-01-07 2019-01-08 2019-01-09 2019-01-10 2019-01-11   

       6          7          8          9  
0 2019-01-07 2019-01-08 2019-01-09 2019-01-10  
1 2019-01-08        NaT        NaT        NaT  
2 2019-01-12 2019-01-13 2019-01-14        NaT