Question

我有一个熊猫数据框：

#define BILLION 1000000000L

#define LIMIT_I 1000
#define LIMIT_J 1000

double get_current_time_seconds1()
{
    /* Get current time using gettimeofday */
    time_t t = time(NULL);
    struct tm *tm = localtime(&t);
    printf("%s\n", asctime(tm));
    return (double) tm;
}

double get_current_time_seconds2()
{
    struct timespec start,stop;
    clock_gettime(CLOCK_REALTIME, &start);
    clock_gettime(CLOCK_REALTIME, &stop);
    double x = (stop.tv_sec - start.tv_sec) + (stop.tv_nsec - start.tv_nsec);

    printf("%lf\n", x);

    return (double) x;
}

double get_current_time_seconds3()
{
    uint64_t diff;

    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);
    sleep(5);
    clock_gettime(CLOCK_MONOTONIC, &end);

    diff = BILLION * (end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
    printf("elapsed time = %llu nanoseconds\n", (long long unsigned int)diff);

    return (double) diff;
}

和一个变量：

我想通过首先按X行数进行分组将其转换为另一个数据框。计算平均值，然后对下一个y行进行分组，计算平均值，依此类推。上面df的结果应该是：

x = 3
y = 2

Answer 1

使用np.put进行检查，然后我们使用diff和cumsum创建组密钥

df['new']=0
np.put(df['new'],np.arange(len(df)),x*[1]+y*[2])
df.groupby(df.new.diff().ne(0).cumsum()).col1.mean()
Out[588]: 
new
1    1
2    2
3    1
4    2
Name: col1, dtype: int64

通过对每X个行数进行分组，然后对下一个Y-行数进行分组来获取均值

1 个答案: