Question

我想计算不包括值99的平均年龄。在现实生活中，数据框要大得多，并且我还有其他可能的变量。

是否有更有效的方法（更快或更优雅）来做到这一点？也许有数据透视表或分组依据，还是有函数？

data = {'age': [99,45,34,32,34,67,5,6,7,8,3,5]}
df = pd.DataFrame(data, columns = ['age'])

not99 = df['age'] != 99

mean_for_age = df.loc[not99, 'age'].mean()

Answer 1

numpy解决方案更快-首先创建数组，然后过滤：

arr = df['age'].values
not99 = arr != 99

mean_for_age = arr[not99].mean()

但是，如果通常需要解决方案以选择另一列，请使用您的解决方案：

not99 = df['age'] != 99
mean_for_age = df.loc[not99, 'age'].mean()

mean_for_age = df.loc[not99, 'another col'].mean()

时间（取决于数据，对真实数据的最佳测试）：

data = {'age': [99,45,34,32,34,67,5,6,7,8,3,5]}
df = pd.DataFrame(data, columns = ['age'])

df = pd.concat([df] * 10000, ignore_index=True)


In [14]: %%timeit
    ...: arr = df['age'].values
    ...: not99 = arr != 99
    ...: 
    ...: mean_for_age = arr[not99].mean()
    ...: 
496 µs ± 36.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [15]: %%timeit
    ...: not99 = df['age'] != 99
    ...: mean_for_age = df.loc[not99, 'age'].mean()
    ...: 
1.82 ms ± 40.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [16]: %%timeit
    ...: df.query("age != 99")['age'].mean()
    ...: 
4.26 ms ± 40.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

熊猫的平均值与where条件

1 个答案: