求Numpy阵列中非均匀元素的指数

Question

我发现了其他方法，比如this，可以从数组中删除重复的元素。我的要求稍有不同。如果我先说：

array([[1, 2, 3],
       [2, 3, 4],
       [1, 2, 3],
       [3, 2, 1],
       [3, 4, 5]])

最后我想说：

^{pr2}$

这就是我最终想要的，但还有一个额外的要求。我还想存储一个要丢弃或保留的索引数组（lanumpy.take）。在

我使用的是Numpy 1.8.1

Answer 1

如果要删除重复版本中存在的元素的所有实例，可以遍历数组，找到多个版本中存在的元素的索引，最后删除这些：

# The array to check:
array = numpy.array([[1, 2, 3],
        [2, 3, 4],
        [1, 2, 3],
        [3, 2, 1],
        [3, 4, 5]])

# List that contains the indices of duplicates (which should be deleted)
deleteIndices = []

for i in range(0,len(array)): # Loop through entire array
    indices = range(0,len(array)) # All indices in array
    del indices[i] # All indices in array, except the i'th element currently being checked

for j in indexes: # Loop through every other element in array, except the i'th element, currently being checked
    if(array[i] == array[j]).all(): # Check if element being checked is equal to the j'th element
        deleteIndices.append(j) # If i'th and j'th element are equal, j is appended to deleteIndices[]

# Sort deleteIndices in ascending order:
deleteIndices.sort()

# Delete duplicates
array = numpy.delete(array,deleteIndices,axis=0)

该输出：

^{pr2}$

这样一来，您既可以删除重复项，又可以获得一个要丢弃的索引列表。在