我找到了其他方法，例如this，可以从数组中删除重复元素。我的要求略有不同。如果我开始：
array([[1, 2, 3],
[2, 3, 4],
[1, 2, 3],
[3, 2, 1],
[3, 4, 5]])
我想结束：
array([[2, 3, 4],
[3, 2, 1]
[3, 4, 5]])
这就是我最终想要的结果，但还有一个额外的要求。我还想存储要丢弃或保留的索引数组（例如numpy.take）。
我正在使用 Numpy 1.8.1
【问题讨论】：
return_counts 所以#1 不适合我。不幸的是＃2似乎需要排序数组，我需要保留顺序。
标签： python arrays python-2.7 numpy
我们希望找到在您的数组中不重复的行，同时保留顺序。
我使用这个solution 将a 的每一行组合成一个元素，这样我们就可以使用np.unique(,return_index=True, return_inverse= True) 找到唯一的行。然后，我修改了这个function 以使用索引和逆输出唯一行的计数。从那里，我可以选择所有具有counts == 1 的唯一行。
a = np.array([[1, 2, 3],
[2, 3, 4],
[1, 2, 3],
[3, 2, 1],
[3, 4, 5]])
#use a flexible data type, np.void, to combine the columns of `a`
#size of np.void is the number of bytes for an element in `a` multiplied by number of columns
b = a.view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))
_, index, inv = np.unique(b, return_index = True, return_inverse = True)
def return_counts(index, inv):
count = np.zeros(len(index), np.int)
np.add.at(count, inv, 1)
return count
counts = return_counts(index, inv)
#if you want the indices to discard replace with: counts[i] > 1
index_keep = [i for i, j in enumerate(index) if counts[i] == 1]
>>>a[index_keep]
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
#if you don't need the indices and just want the array returned while preserving the order
a_unique = np.vstack(a[idx] for i, idx in enumerate(index) if counts[i] == 1])
>>>a_unique
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
对于 np.version >= 1.9
b = a.view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))
_, index, counts = np.unique(b, return_index = True, return_counts = True)
index_keep = [i for i, j in enumerate(index) if counts[i] == 1]
>>>a[index_keep]
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
【讨论】：
您可以按照以下方式进行：
# Assuming your array is a
uniq, uniq_idx, counts = np.unique(a, axis=0, return_index=True, return_counts=True)
# to return the array you want
new_arr = uniq[counts == 1]
# The indices of non-unique rows
a_idx = np.arange(a.shape[0]) # the indices of array a
nuniq_idx = a_idx[np.in1d(a_idx, uniq_idx[counts==1], invert=True)]
你得到：
#new_arr
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
# nuniq_idx
array([0, 2])
【讨论】：
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)
这个输出：
>>> array
array([[2, 3, 4],
[3, 2, 1],
[3, 4, 5]])
>>> deleteIndices
[0, 2]
就像您删除重复项并获得要丢弃的索引列表一样。
【讨论】：
numpy_indexed 包（免责声明：我是它的作者）可用于以矢量化方式解决此类问题：
index = npi.as_index(arr)
keep = index.count == 1
discard = np.invert(keep)
print(index.unique[keep])
【讨论】：