Question

我有以下数据框：

 public static boolean isAvailable(Context context)
    {
        GoogleApiAvailability availability = GoogleApiAvailability.getInstance();

        return isGooglePlayServicesAvailable(context, availability) &&
                isCastContextAvailable(context);
    }

public static boolean isAvailable(Context context) {
  if (googlePlayServicesVerified(context)) {
    try {
      castContext = CastContext.getSharedInstance(context);
      Log.d(TAG, "CastContext instantiated");
    } catch (Exception e) {
      Log.report(e);
      castContext = null;
    }
  } else {
    CrashReporter.report("CastContext FAILED to be instantiated : googlePlayServicesVerified() has failed."));
    castContext = null;
  }
}

我想用time = 2填充时间戳中的所有空白，因此输出将如下：

    timestamp            value
    2018-02-26 09:13:00  0.972198
    2018-02-26 09:14:00  1.008504
    2018-02-26 09:15:00  1.011961
    2018-02-26 09:18:00  1.018950
    2018-02-26 09:19:00  1.008538
    2018-02-26 09:21:00  0.988535
    2018-02-26 09:22:00  0.944170
    2018-02-26 09:23:00  0.940284

我使用以下代码首先填补时间戳中的空白：

    timestamp            value
    2018-02-26 09:13:00  0.972198
    2018-02-26 09:14:00  1.008504
    2018-02-26 09:15:00  1.011961
    2018-02-26 09:16:00  2.0
    2018-02-26 09:17:00  2.0
    2018-02-26 09:18:00  1.018950
    2018-02-26 09:19:00  1.008538
    2018-02-26 09:20:00  2.0
    2018-02-26 09:21:00  0.988535
    2018-02-26 09:22:00  0.944170
    2018-02-26 09:23:00  0.940284

但出现以下错误。我想知道我在这里想念的是什么？谢谢！

df.reindex(index = 'timestamp')

Answer 1

使用resample + fillna

df.set_index('timestamp').resample('60s').mean().fillna(2).reset_index()
Out[907]: 
             timestamp     value
0  2018-02-26 09:13:00  0.972198
1  2018-02-26 09:14:00  1.008504
2  2018-02-26 09:15:00  1.011961
3  2018-02-26 09:16:00  2.000000
4  2018-02-26 09:17:00  2.000000
5  2018-02-26 09:18:00  1.018950
6  2018-02-26 09:19:00  1.008538
7  2018-02-26 09:20:00  2.000000
8  2018-02-26 09:21:00  0.988535
9  2018-02-26 09:22:00  0.944170
10 2018-02-26 09:23:00  0.940284

Answer 2

date_range

ts = pd.date_range(df.timestamp.min(), df.timestamp.max(), freq='1min')

set_index ， reindex 和 fillna

df.set_index('timestamp').reindex(ts).fillna(2.0).rename_axis('timestamp').reset_index()

             timestamp     value
0  2018-02-26 09:13:00  0.972198
1  2018-02-26 09:14:00  1.008504
2  2018-02-26 09:15:00  1.011961
3  2018-02-26 09:16:00  2.000000
4  2018-02-26 09:17:00  2.000000
5  2018-02-26 09:18:00  1.018950
6  2018-02-26 09:19:00  1.008538
7  2018-02-26 09:20:00  2.000000
8  2018-02-26 09:21:00  0.988535
9  2018-02-26 09:22:00  0.944170
10 2018-02-26 09:23:00  0.940284

熊猫：以一定的价值及时填补空白

2 个答案: