如何计算numpy中多个项目的出现次数？

Question

假设以下numpy数组：

[1,2,3,1,2,3,1,2,3,1,2,2]

我想count([1,2])一次计算所有出现的1和2，得出类似

[4, 5]

对应于[1, 2]输入。

它在numpy中受支持吗？

Answer 1

# Setting your input to an array
array = np.array([1,2,3,1,2,3,1,2,3,1,2,2])

# Find the unique elements and get their counts
unique, counts = np.unique(array, return_counts=True)

# Setting the numbers to get counts for as an array
search = np.array([1, 2])

# Gets the counts for the elements in search
search_counts = [counts[i] for i, x in enumerate(unique) if x in search]

这将输出[4，5]