Question

我想计算从数据集中停止服务器的时间长度。我知道停机时间，但不知道停机时间。

我有这个df：

index                   a          b     c     reboot
2018-06-25 12:51:00    NaN        NaN   NaN     1      
2018-06-25 12:52:00    NaN        NaN   NaN     0    
2018-06-25 12:53:00    NaN        NaN   NaN     0  
2018-06-25 12:54:00    NaN        NaN   NaN     0    
2018-06-25 12:55:00    NaN        NaN   NaN     0    
2018-06-25 12:56:00    NaN        NaN   NaN     0   
2018-06-25 12:57:00    NaN        NaN   NaN     0   
2018-06-25 12:58:00    NaN        0.6   0.6     0
2018-06-25 12:59:00    NaN        NaN   0.5     0  
2018-06-25 13:00:00    NaN        NaN   0.3     0  
2018-06-25 13:01:00   2.55  94.879997  0.23     0
2018-06-25 13:02:00   1.17        Nan  0.13     0
2018-06-25 13:03:00   1.08  98.199997  0.10     0
2018-06-25 13:28:00    NaN        NaN   NaN     1  
2018-06-25 13:29:00    NaN        NaN   NaN     0     
2018-06-25 13:30:00    NaN        NaN   NaN     0
2018-06-25 13:31:00    NaN        NaN   NaN     0
2018-06-25 13:31:00    0.5        0.2   0.1     0
2018-06-25 13:32:00    NaN        NaN   NaN     0 
2018-06-25 13:33:00    NaN        NaN   NaN     0 
2018-06-25 13:34:00     3         0.6   0.5     0

我要计算a，b和c分别为NaN和reboot == 1的行，其结果采用以下形式：

index                    period      reboot
2018-06-25 12:51:00         7           1
2018-06-25 13:28:00         4           1

我已经尝试了在没有重新启动条件的情况下逐列进行操作。

输入：

index                   a          b     c     reboot
2018-06-25 12:51:00    NaN        NaN   NaN     1      
2018-06-25 12:52:00    NaN        NaN   NaN     0    
2018-06-25 12:53:00    NaN        NaN   NaN     0  
2018-06-25 12:54:00    NaN        NaN   NaN     0    
2018-06-25 12:55:00    NaN        NaN   NaN     0    
2018-06-25 12:56:00    NaN        NaN   NaN     0   
2018-06-25 12:57:00    NaN        NaN   NaN     0   
2018-06-25 12:58:00    NaN        NaN   NaN     0
2018-06-25 12:59:00    NaN        NaN   NaN     0  
2018-06-25 13:00:00    NaN        NaN   NaN     0  
2018-06-25 13:01:00   2.55  94.879997  0.23     0
2018-06-25 13:02:00   1.17        Nan  0.13     0
2018-06-25 13:03:00   1.08  98.199997  0.10     0
2018-06-25 13:28:00    NaN        NaN   NaN     1  
2018-06-25 13:29:00    NaN        NaN   NaN     0     
2018-06-25 13:30:00    NaN        NaN   NaN     0

a=df.index
b=df.b.values
idx0 = np.flatnonzero(np.r_[True, np.diff(np.isnan(b))!=0,True])
count = np.diff(idx0)
idx = idx0[:-1]
valid_mask = (count>=step) & np.isnan(b[idx])
out_idx = idx[valid_mask]
out_num = a[out_idx]
out_count = count[valid_mask]
outb = zip(out_num, out_count)
periodb=list(outb)

结果：

'[(Timestamp('2018-06-25 12:51:00'), 10),
 (Timestamp('2018-06-25 13:28:00'), 3),'

Answer 1

Add的另一列具有“正常”索引（整数从0开始计数），选择感兴趣的行，然后选择find the differences between adjacent values in the added column-因为这些差异将为您提供原始行之间的距离数据。

类似的东西：

numbered = df.assign(row=range(len(df)))
restarts = numbered[numbered.reboot == 1]
result = restarts.row.shift(-1) - restarts.row

（仔细一点看，问题的一部分似乎是只对所有a，b，c值都用NaN计算行。为此，请过滤掉所有 other first 行，然后添加辅助索引列。）

使用条件对更多列中的连续缺失值（nan）进行计数

1 个答案: