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pandas dataframe-具有选定开始字符串和指定列的列

pandas

tandem · 技术社区 · 6 年前

在下面给定的数据帧中,我想清理计数器(例如MEM\u TRANS\u RETIRED)并将列添加到数据帧中 load,rps,th95

       load        rps     th95    energy     ...       MEM_TRANS_RETIRED-34  PERF_COUNT_HW_CACHE_L1D-34  PERF_COUNT_HW_CACHE_L1I-34     map_freq
0    500.0k  346222.62    12.62   7270.22     ...                  154287.14                   591053.74                2.918521e+07   6C-1.70GHz
1    400.0k  402628.34     2.25  12026.40     ...                  189915.07                   627043.91                2.867945e+07  10C-2.10GHz
2    500.0k  283508.27    15.52   5662.74     ...                  140790.31                  1431892.98                4.253950e+07   6C-1.30GHz

我就是这样做的

self.unique_counters = [x[:-2] for x in self.dfile_keys[6:] if x.endswith('-0')]
for counter in self.unique_counters:
 new = self.dfile.loc[:, self.dfile.columns.str.startswith(counter)]

     PERF_COUNT_HW_CPU_CYCLES-0  PERF_COUNT_HW_CPU_CYCLES-2             ...               PERF_COUNT_HW_CPU_CYCLES-32  PERF_COUNT_HW_CPU_CYCLES-34
0                  6.020913e+08                6.021277e+08             ...                              5.109342e+06                 2.556039e+06
1                  4.781879e+08                4.783621e+08             ...                              3.095814e+06                 2.795868e+06
2                  4.841784e+08                4.844846e+08             ...                              2.389396e+06                 5.550159e+06

如何获得具有选定开始字符串和某些指定列的计数器这是预期的输出

       load        rps     th95    energy   PERF_COUNT_HW_CPU_CYCLES-0  PERF_COUNT_HW_CPU_CYCLES-2             ...               PERF_COUNT_HW_CPU_CYCLES-32  PERF_COUNT_HW_CPU_CYCLES-34
0    500.0k  346222.62    12.62   7270.22   6.020913e+08                6.021277e+08             ...                              5.109342e+06                 2.556039e+06
1    400.0k  402628.34     2.25  12026.40   4.781879e+08                4.783621e+08             ...                              3.095814e+06                 2.795868e+06
2    500.0k  283508.27    15.52   5662.74   4.841784e+08                4.844846e+08             ...                              2.389396e+06                 5.550159e+06

1 回复 | 直到 6 年前

jezrael 6 年前

我相信你需要一个新的 DataFrame :

L = [x[:-2] for x in self.dfile_keys[6:] if x.endswith('-0')]

new = self.dfile.loc[:, self.dfile.columns.str.startswith(tuple(L))]

df = pd.concat([self.dfile[['load','rps','th95','energy']], new], axis=1)

list of DataFrame 学生:

self.unique_counters = [x[:-2] for x in self.dfile_keys[6:] if x.endswith('-0')]
dfs = [self.dfile.loc[:, self.dfile.columns.str.startswith(counter)] 
       for counter in self.unique_counters]

df = pd.concat([self.dfile[['load','rps','th95','energy']], dfs], axis=1)