将相同的词典顺序分配给二维数组的重复元素
Assign same lexicographic rank to duplicate elements of 2d array
我正在尝试按字典顺序排列数组组件。下面的代码工作正常,但我想为相等的元素分配相等的等级。
import numpy as np
values = np.asarray([
[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]
])
# need to flip, because for `np.lexsort` last
# element has highest priority.
values_reversed = np.fliplr(values)
# this returns the order, i.e. the order in
# which the elements should be in a sorted
# array (not the rank by index).
order = np.lexsort(values_reversed.T)
# convert order to ranks.
n = values.shape[0]
ranks = np.empty(n, dtype=int)
# use order to assign ranks.
ranks[order] = np.arange(n)
等级变量包含 [2, 0, 4, 3, 1]
,但需要 [2, 0, 4, 2, 1]
的等级数组,因为元素 [1, 2, 3]
(索引 0 和 3)共享相同的等级。连续的排名数字是可以的,所以 [2, 0, 3, 2, 1]
也是一个可接受的排名数组。
这是一种方法 -
# Get lexsorted indices and hence sorted values by those indices
lexsort_idx = np.lexsort(values.T[::-1])
lexsort_vals = values[lexsort_idx]
# Mask of steps where rows shift (there are no duplicates in subsequent rows)
mask = np.r_[True,(lexsort_vals[1:] != lexsort_vals[:-1]).any(1)]
# Get the stepped indices (indices shift at non duplicate rows) and
# the index values are scaled corresponding to row numbers
stepped_idx = np.maximum.accumulate(mask*np.arange(mask.size))
# Re-arrange the stepped indices based on the original order of rows
# This is basically same as the original code does in last 4 steps,
# just in a concise manner
out_idx = stepped_idx[lexsort_idx.argsort()]
逐步中间输出示例 -
In [55]: values
Out[55]:
array([[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]])
In [56]: lexsort_idx
Out[56]: array([1, 4, 0, 3, 2])
In [57]: lexsort_vals
Out[57]:
array([[1, 1, 1],
[1, 1, 2],
[1, 2, 3],
[1, 2, 3],
[2, 2, 3]])
In [58]: mask
Out[58]: array([ True, True, True, False, True], dtype=bool)
In [59]: stepped_idx
Out[59]: array([0, 1, 2, 2, 4])
In [60]: lexsort_idx.argsort()
Out[60]: array([2, 0, 4, 3, 1])
In [61]: stepped_idx[lexsort_idx.argsort()]
Out[61]: array([2, 0, 4, 2, 1])
性能提升
为了提高计算 lexsort_idx.argsort()
的性能效率,我们可以使用这与最后 4 行中的原始代码相同 -
def argsort_unique(idx):
# Original idea : by @Andras
n = idx.size
sidx = np.empty(n,dtype=int)
sidx[idx] = np.arange(n)
return sidx
因此,lexsort_idx.argsort()
可以用 argsort_unique(lexsort_idx)
替代计算。
运行时测试
再应用一些优化技巧,我们就会有一个像这样的版本 -
def numpy_app(values):
lexsort_idx = np.lexsort(values.T[::-1])
lexsort_v = values[lexsort_idx]
mask = np.concatenate(( [False],(lexsort_v[1:] == lexsort_v[:-1]).all(1) ))
stepped_idx = np.arange(mask.size)
stepped_idx[mask] = 0
np.maximum.accumulate(stepped_idx, out=stepped_idx)
return stepped_idx[argsort_unique(lexsort_idx)]
@Warren Weckesser 基于排名数据的方法作为计时函数 -
def scipy_app(values):
v = values.view(np.dtype(','.join([values.dtype.str]*values.shape[1])))
return rankdata(v, method='min') - 1
计时 -
In [97]: a = np.random.randint(0,9,(10000,3))
In [98]: out1 = numpy_app(a)
In [99]: out2 = scipy_app(a)
In [100]: np.allclose(out1, out2)
Out[100]: True
In [101]: %timeit scipy_app(a)
100 loops, best of 3: 5.32 ms per loop
In [102]: %timeit numpy_app(a)
100 loops, best of 3: 1.96 ms per loop
这是一种使用 scipy.stats.rankdata
(使用 method='min'
)的方法,方法是将二维数组视为一维结构化数组:
In [15]: values
Out[15]:
array([[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]])
In [16]: v = values.view(np.dtype(','.join([values.dtype.str]*values.shape[1])))
In [17]: rankdata(v, method='min') - 1
Out[17]: array([2, 0, 4, 2, 1])
我正在尝试按字典顺序排列数组组件。下面的代码工作正常,但我想为相等的元素分配相等的等级。
import numpy as np
values = np.asarray([
[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]
])
# need to flip, because for `np.lexsort` last
# element has highest priority.
values_reversed = np.fliplr(values)
# this returns the order, i.e. the order in
# which the elements should be in a sorted
# array (not the rank by index).
order = np.lexsort(values_reversed.T)
# convert order to ranks.
n = values.shape[0]
ranks = np.empty(n, dtype=int)
# use order to assign ranks.
ranks[order] = np.arange(n)
等级变量包含 [2, 0, 4, 3, 1]
,但需要 [2, 0, 4, 2, 1]
的等级数组,因为元素 [1, 2, 3]
(索引 0 和 3)共享相同的等级。连续的排名数字是可以的,所以 [2, 0, 3, 2, 1]
也是一个可接受的排名数组。
这是一种方法 -
# Get lexsorted indices and hence sorted values by those indices
lexsort_idx = np.lexsort(values.T[::-1])
lexsort_vals = values[lexsort_idx]
# Mask of steps where rows shift (there are no duplicates in subsequent rows)
mask = np.r_[True,(lexsort_vals[1:] != lexsort_vals[:-1]).any(1)]
# Get the stepped indices (indices shift at non duplicate rows) and
# the index values are scaled corresponding to row numbers
stepped_idx = np.maximum.accumulate(mask*np.arange(mask.size))
# Re-arrange the stepped indices based on the original order of rows
# This is basically same as the original code does in last 4 steps,
# just in a concise manner
out_idx = stepped_idx[lexsort_idx.argsort()]
逐步中间输出示例 -
In [55]: values
Out[55]:
array([[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]])
In [56]: lexsort_idx
Out[56]: array([1, 4, 0, 3, 2])
In [57]: lexsort_vals
Out[57]:
array([[1, 1, 1],
[1, 1, 2],
[1, 2, 3],
[1, 2, 3],
[2, 2, 3]])
In [58]: mask
Out[58]: array([ True, True, True, False, True], dtype=bool)
In [59]: stepped_idx
Out[59]: array([0, 1, 2, 2, 4])
In [60]: lexsort_idx.argsort()
Out[60]: array([2, 0, 4, 3, 1])
In [61]: stepped_idx[lexsort_idx.argsort()]
Out[61]: array([2, 0, 4, 2, 1])
性能提升
为了提高计算 lexsort_idx.argsort()
的性能效率,我们可以使用这与最后 4 行中的原始代码相同 -
def argsort_unique(idx):
# Original idea : by @Andras
n = idx.size
sidx = np.empty(n,dtype=int)
sidx[idx] = np.arange(n)
return sidx
因此,lexsort_idx.argsort()
可以用 argsort_unique(lexsort_idx)
替代计算。
运行时测试
再应用一些优化技巧,我们就会有一个像这样的版本 -
def numpy_app(values):
lexsort_idx = np.lexsort(values.T[::-1])
lexsort_v = values[lexsort_idx]
mask = np.concatenate(( [False],(lexsort_v[1:] == lexsort_v[:-1]).all(1) ))
stepped_idx = np.arange(mask.size)
stepped_idx[mask] = 0
np.maximum.accumulate(stepped_idx, out=stepped_idx)
return stepped_idx[argsort_unique(lexsort_idx)]
@Warren Weckesser 基于排名数据的方法作为计时函数 -
def scipy_app(values):
v = values.view(np.dtype(','.join([values.dtype.str]*values.shape[1])))
return rankdata(v, method='min') - 1
计时 -
In [97]: a = np.random.randint(0,9,(10000,3))
In [98]: out1 = numpy_app(a)
In [99]: out2 = scipy_app(a)
In [100]: np.allclose(out1, out2)
Out[100]: True
In [101]: %timeit scipy_app(a)
100 loops, best of 3: 5.32 ms per loop
In [102]: %timeit numpy_app(a)
100 loops, best of 3: 1.96 ms per loop
这是一种使用 scipy.stats.rankdata
(使用 method='min'
)的方法,方法是将二维数组视为一维结构化数组:
In [15]: values
Out[15]:
array([[1, 2, 3],
[1, 1, 1],
[2, 2, 3],
[1, 2, 3],
[1, 1, 2]])
In [16]: v = values.view(np.dtype(','.join([values.dtype.str]*values.shape[1])))
In [17]: rankdata(v, method='min') - 1
Out[17]: array([2, 0, 4, 2, 1])