Question

假设我有一个numpy数组，它在两种项目类型的ID之间进行映射：

[[1, 12],
 [1, 13],
 [1, 14],
 [2, 13],
 [2, 14],
 [3, 11]]

我想重新排列此数组，以便新数组中的每一行代表与原始数组中相同ID匹配的所有项。这里，每列代表原始数组中的一个映射，直到新数组中列数的指定形状限制。如果我们想从上面的数组中获得这个结果，确保我们只有2列，我们将获得：

[[12, 13],  #Represents 1 - 14 was not kept as only 2 columns are allowed
 [13, 14],  #Represents 2
 [11,  0]]  #Represents 3 - 0 was used as padding since 3 did not have 2 mappings

这里天真的方法是使用for循环来填充新数组，因为它遇到原始数组中的行。是否有更有效的方法来实现numpy的功能？

Answer 1

这是一般的，主要是Numpythonic方法：

In [144]: def array_packer(arr):
     ...:     cols = arr.shape[1]
     ...:     ids = arr[:, 0]
     ...:     inds = np.where(np.diff(ids) != 0)[0] + 1
     ...:     sp = np.split(arr[:,1:], inds)
     ...:     result = [np.unique(a[: cols]) if a.shape[0] >= cols else
     ...:                    np.pad(np.unique(a), (0, (cols - 1) * (cols - a.shape[0])), 'constant')
     ...:                 for a in sp]
     ...:     return result
     ...:     
     ...:

演示：

In [145]: a = np.array([[1, 12, 15, 45],
     ...:  [1, 13, 23, 9],
     ...:  [1, 14, 14, 11],
     ...:  [2, 13, 90, 34],
     ...:  [2, 14, 23, 43],
     ...:  [3, 11, 123, 53]])
     ...:  

In [146]: array_packer(a)
Out[146]: 
[array([ 9, 11, 12, 13, 14, 15, 23, 45,  0,  0,  0]),
 array([13, 14, 23, 34, 43, 90,  0,  0,  0,  0,  0,  0]),
 array([ 11,  53, 123,   0,   0,   0,   0,   0,   0,   0,   0,   0])]

In [147]: a = np.array([[1, 12, 15],
     ...:  [1, 13, 23],
     ...:  [1, 14, 14],
     ...:  [2, 13, 90],
     ...:  [2, 14, 23],
     ...:  [3, 11, 123]])
     ...: 
     ...:   
     ...:  

In [148]: array_packer(a)
Out[148]: 
[array([12, 13, 14, 15, 23]),
 array([13, 14, 23, 90,  0,  0]),
 array([ 11, 123,   0,   0,   0,   0])]

Answer 2

对于这个问题，天真的for循环实际上是一个非常有效的解决方案：

from collections import defaultdict, deque
d = defaultdict(lambda: deque((0, 0), maxlen=2))

%%timeit
for key, val in a:
    d[key].append(val)
4.43 µs ± 29.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

# result: {1: deque([13, 14]), 2: deque([13, 14]), 3: deque([0, 11])}

为了进行比较，这个线程中提出的numpy解决方案慢了4倍：

%timeit [[*a[a[:,0]==i,1],0][:2] for i in np.unique(a[:,0])]
18.6 µs ± 336 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Numpy非常好，我自己也经常使用它，但我认为这种情况很麻烦。

Answer 3

这是一种使用稀疏矩阵的方法：

def pp(map_, maxitems=2):
    M = sparse.csr_matrix((map_[:, 1], map_[:, 0], np.arange(map_.shape[0]+1)))
    M = M.tocsc()
    sizes = np.diff(M.indptr)
    ids, = np.where(sizes)
    D = np.concatenate([M.data, np.zeros((maxitems - 1,), dtype=M.data.dtype)])
    D = np.lib.stride_tricks.as_strided(D, (D.size - maxitems + 1, maxitems),
                                        2 * D.strides)
    result = D[M.indptr[ids]]
    result[np.arange(maxitems) >= sizes[ids, None]] = 0
    return result

使用@crisz的代码进行计时，但修改后使用较少重复的测试数据。我还添加了一些“验证”：chrisz和我的解决方案给出相同的答案，另外两个输出不同的格式，所以我无法检查它们。

代码：

from scipy import sparse
import numpy as np
from collections import defaultdict, deque

def pp(map_, maxitems=2):
    M = sparse.csr_matrix((map_[:, 1], map_[:, 0], np.arange(map_.shape[0]+1)))
    M = M.tocsc()
    sizes = np.diff(M.indptr)
    ids, = np.where(sizes)
    D = np.concatenate([M.data, np.zeros((maxitems - 1,), dtype=M.data.dtype)])
    D = np.lib.stride_tricks.as_strided(D, (D.size - maxitems + 1, maxitems),
                                        2 * D.strides)
    result = D[M.indptr[ids]]
    result[np.arange(maxitems) >= sizes[ids, None]] = 0
    return result

def chrisz(a):
  return [[*a[a[:,0]==i,1],0][:2] for i in np.unique(a[:,0])]

def piotr(a):
  d = defaultdict(lambda: deque((0, 0), maxlen=2))
  for key, val in a:
    d[key].append(val)
  return d

def karams(arr):
  cols = arr.shape[1]
  ids = arr[:, 0]
  inds = np.where(np.diff(ids) != 0)[0] + 1
  sp = np.split(arr[:,1:], inds)
  result = [a[:2].ravel() if a.size >= cols else np.pad(a.ravel(), (0, cols -1 * (cols - a.size)), 'constant')for a in sp]
  return result

def make(nid, ntot):
    return np.c_[np.random.randint(0, nid, (ntot,)),
                 np.random.randint(0, 2**30, (ntot,))]

from timeit import timeit
import pandas as pd
import matplotlib.pyplot as plt

res = pd.DataFrame(
       index=['pp', 'chrisz', 'piotr', 'karams'],
       columns=[10, 50, 100, 500, 1000, 5000, 10000],# 50000],
       dtype=float
)

for c in res.columns:
#        l = np.repeat(np.array([[1, 12],[1, 13],[1, 14],[2, 13],[2, 14],[3, 11]]), c, axis=0)
    l = make(c // 2, c * 6)
    assert np.all(chrisz(l) == pp(l))
    for f in res.index:
        stmt = '{}(l)'.format(f)
        setp = 'from __main__ import l, {}'.format(f)
        res.at[f, c] = timeit(stmt, setp, number=30)

ax = res.div(res.min()).T.plot(loglog=True)
ax.set_xlabel("N");
ax.set_ylabel("time (relative)");

plt.show()

Answer 4

略微改编自几乎复制到pad并仅选择两个元素：

[[*a[a[:,0]==i,1],0][:2] for i in np.unique(a[:,0])]

输出：

[[12, 13], [13, 14], [11, 0]]

如果您想跟踪密钥：

{i:[*a[a[:,0]==i,1],0][:2] for i in np.unique(a[:,0])}

# {1: [12, 13], 2: [13, 14], 3: [11, 0]}

<强> 功能

def chrisz(a):
  return [[*a[a[:,0]==i,1],0][:2] for i in np.unique(a[:,0])]

def piotr(a):
  d = defaultdict(lambda: deque((0, 0), maxlen=2))
  for key, val in a:
    d[key].append(val)
  return d

def karams(arr):
  cols = arr.shape[1]
  ids = arr[:, 0]
  inds = np.where(np.diff(ids) != 0)[0] + 1
  sp = np.split(arr[:,1:], inds)
  result = [a[:2].ravel() if a.size >= cols else np.pad(a.ravel(), (0, cols -1 * (cols - a.size)), 'constant')for a in sp]
  return result

<强> 计时

from timeit import timeit
import pandas as pd
import matplotlib.pyplot as plt

res = pd.DataFrame(
       index=['chrisz', 'piotr', 'karams'],
       columns=[10, 50, 100, 500, 1000, 5000, 10000, 50000],
       dtype=float
)

for f in res.index:
    for c i

n res.columns:
        l = np.repeat(np.array([[1, 12],[1, 13],[1, 14],[2, 13],[2, 14],[3, 11]]), c, axis=0)
        stmt = '{}(l)'.format(f)
        setp = 'from __main__ import l, {}'.format(f)
        res.at[f, c] = timeit(stmt, setp, number=30)

ax = res.div(res.min()).T.plot(loglog=True)
ax.set_xlabel("N");
ax.set_ylabel("time (relative)");

plt.show()

结果（显然@Kasramvd是赢家）：

根据ID将numpy行转换为列

4 个答案: