@Divakar刚刚发布了一个很好的答案.如果您已经定义了类别数组，则可以使用@Divakar的答案.如果您尚未定义唯一值，则使用我的.

@Divakar just posted a very good answer. If you already have the array of categories defined, I'd use @Divakar's answer. If you don't have unique values already define, I'd use mine.

我将使用 pd.factorize 分解类别.然后使用 np.bincount 参数weights设置为values数组的a>

I'd use pd.factorize to factorize the categories. Then use np.bincount with weights parameter set to be the values array

f, u = pd.factorize(valcats)
np.bincount(f, values).astype(values.dtype)

array([ 1, 12,  7, 14, 13,  8])

pd.factorize还会在u变量中产生唯一值.我们可以将结果与u对齐，以查看是否找到了正确的解决方案.

pd.factorize also produces the unique values in the u variable. We can line up the results with u to see that we've arrived at the correct solution.

np.column_stack([u, np.bincount(f, values).astype(values.dtype)])

array([[101,   1],
       [301,  12],
       [201,   7],
       [102,  14],
       [302,  13],
       [202,   8]])

您可以使用pd.Series

f, u = pd.factorize(valcats)
pd.Series(np.bincount(f, values).astype(values.dtype), u)

101     1
301    12
201     7
102    14
302    13
202     8
dtype: int64

为什么 pd.factorize 而不是 np.unique ?

Why pd.factorize and not np.unique?

我们本可以用

 u, f = np.unique(valcats, return_inverse=True)

但是，np.unique对值进行排序，并在nlogn时间运行.另一方面，pd.factorize不排序，并且在线性时间内运行.对于较大的数据集，pd.factorize将主导性能.

But, np.unique sorts the values and that runs in nlogn time. On the other hand pd.factorize does not sort and runs in linear time. For larger data sets, pd.factorize will dominate performance.