python - Pandas 蟒 : how to get values on data frame every 10 steps?

Question

我有一个简单的问题。我有以下数据框

df =
    time                                        lat          lon
    0   2014-03-26 14:46:27.457233+00:00    48.7773     11.428897
    1   2014-03-26 14:46:28.457570+00:00    48.7773     11.428719
    2   2014-03-26 14:46:29.457665+00:00    48.7772     11.428542
    3   2014-03-26 14:46:30.457519+00:00    48.7771     11.428368
    4   2014-03-26 14:46:31.457855+00:00    48.7770     11.428193
    5   2014-03-26 14:46:32.457950+00:00    48.7770     11.428018
    6   2014-03-26 14:46:33.457794+00:00    48.7769     11.427842
    7   2014-03-26 14:46:34.458131+00:00    48.7768     11.427668
    8   2014-03-26 14:46:35.458246+00:00    48.7767     11.427501
    9   2014-03-26 14:46:36.458069+00:00    48.7766     11.427350
    10  2014-03-26 14:46:37.458416+00:00    48.7766     11.427224
    11  2014-03-26 14:46:38.458531+00:00    48.7765     11.427129
    12  2014-03-26 14:46:39.458355+00:00    48.7764     11.427062
    13  2014-03-26 14:46:40.458702+00:00    48.7764     11.427011
    14  2014-03-26 14:46:41.458807+00:00    48.7764     11.426963
    15  2014-03-26 14:46:42.458640+00:00    48.7763     11.426918
    16  2014-03-26 14:46:43.458977+00:00    48.7763     11.426872
    17  2014-03-26 14:46:44.459102+00:00    48.7762     11.426822
    18  2014-03-26 14:46:45.458926+00:00    48.7762     11.426766
    19  2014-03-26 14:46:46.459262+00:00    48.7761     11.426702
    20  2014-03-26 14:46:47.459378+00:00    48.7760     11.426628

我想生成一个新的数据帧 df1，其中包含每 10 个时间步长的值。

df1 =
        time                                        lat          lon
        0       2014-03-26 14:46:27.457233+00:00    48.7773     11.428897
        9      2014-03-26 14:46:46.459262+00:00     48.7761     11.426702
        19      2014-03-26 14:46:46.459262+00:00    48.7765     11.426787
        ...        ...         ...                 ...       ....
        len(df) 2014-03-26 14:46:46.459262+00:00    48.7765     11.426787

我试着做类似的事情

df1 = df.iloc[[0:10:len(df)]]

Answer 1

只需使用 iloc 对 df 进行切片并传递一个步骤参数，就可以解释切片行为 here但基本上第三个参数是步长:

In [67]:
df = pd.DataFrame(np.random.randn(100,2))
df.iloc[::10]

Out[67]:
           0         1
0   0.552160 -0.910893
10 -2.173707 -0.659227
20  0.811937  0.675416
30  0.533533  0.336104
40  1.093083 -0.943157
50 -0.559221  0.272763
60 -0.011628  1.002561
70 -0.114501  0.457626
80  1.355948  0.236342
90 -0.151979 -0.746238