从二维 numpy 数组中删除特定行值数组的快速方法

Question

我有一个像这样的二维数组：

a = np.array([[25, 83, 18, 71],
       [75,  7,  0, 85],
       [25, 83, 18, 71],
       [25, 83, 18, 71],
       [75, 48,  8, 43],
       [ 7, 47, 96, 94],
       [ 7, 47, 96, 94],
       [56, 75, 50,  0],
       [19, 49, 92, 57],
       [52, 93, 58,  9]])

我想删除具有特定值的行，例如：

b = np.array([[56, 75, 50,  0], [52, 93, 58,  9], [25, 83, 18, 71]])

在numpy或 中执行此操作的最有效方法是什么pandas？预期输出：

np.array([[75,  7,  0, 85],
       [75, 48,  8, 43],
       [ 7, 47, 96, 94],
       [ 7, 47, 96, 94],
       [19, 49, 92, 57]])

更新

最快的方法是降维，但它通常需要对列的范围进行非常严格的限制。有我的性能图：

import pandas as pd
import numexpr as ne
import perfplot
from time import time

def remove_pd(data):
    a,b = data
    dfa, dfb = pd.DataFrame(a), pd.DataFrame(b)
    return dfa.merge(dfb, how='left', indicator=True)\
    .query('_merge == "left_only"').drop(columns='_merge').values
    
def remove_smalldata(data):
    a,b = data
    return a[(a[None,:,:] != b[:,None,:]).any(-1).all(0)]

'''def remove_nploop(data):
    a, b = data
    for arr in b:
        a = a[np.all(~np.equal(a, arr), axis=1)]
    return a'''
        
def remove_looped(data): 
    a, b = data
    to_remain = [True]*len(a)
    ind = 0
    for vec_a in a:
        for vec_b in b:
            if np.array_equal(vec_a, vec_b):
                to_remain[ind] = False
                break
        ind += 1
    return a[to_remain]

def remove_looped_boost(data): 
    a, b = data
    to_remain = [True]*len(a)
    a_map = list(map(tuple, a.tolist()))
    b_map = set(map(tuple, b.tolist()))
    for i in range(len(a)):
        to_remain[i] = not(a_map[i] in b_map)
    return a[to_remain]

def remove_reducedim(data):
    a,b = data
    a, b = a.astype(np.int64), b.astype(np.int64) #make sure box is not too small
    ma, MA = np.min(a, axis=0), np.max(a, axis=0)
    mb, MB = np.min(b, axis=0), np.max(b, axis=0)
    m, M = np.min([ma, mb], axis=0), np.max([MA, MB],axis=0)
    ravel_a = np.ravel_multi_index((a-m).T, M - m + 1)
    ravel_b = np.ravel_multi_index((b-m).T, M - m + 1)
    return a[~np.isin(ravel_a, ravel_b)]

def remove_reducedim_boost(data):
    a,b = data
    a, b = a.astype(np.int64), b.astype(np.int64) #make sure box is not too small
    ma, MA = np.min(a, axis=0), np.max(a, axis=0)
    mb, MB = np.min(b, axis=0), np.max(b, axis=0)
    m1,m2,m3,m4 = np.min([ma, mb], axis=0)
    M1,M2,M3,M4 = np.max([MA, MB], axis=0)
    s1,s2,s3,s4 = M1-m1+1, M2-m2+1, M3-m3+1, M4-m4+1
    a1,a2,a3,a4 = a.T
    b1,b2,b3,b4 = b.T
    d = {'a1':a1, 'a2':a2, 'a3':a3, 'a4':a4, 'b1':b1, 'b2':b2, 'b3':b3, 'b4':b4,
        's1':s1, 's2':s2, 's3':s3, 'm1':m1, 'm2':m2, 'm3':m3, 'm4':m4}
    ravel_a = ne.evaluate('(a1-m1)+(a2-m2)*s1+(a3-m3)*s1*s2+(a4-m4)*s1*s2*s3',d)
    ravel_b = ne.evaluate('(b1-m1)+(b2-m2)*s1+(b3-m3)*s1*s2+(b4-m4)*s1*s2*s3',d)
    return a[~np.isin(ravel_a, ravel_b)]
    
def setup(x):
    a1 = np.random.randint(50000, size=(x,4))
    a2 = a1[np.random.randint(x, size=x)]
    return a1, a2
    
def build_args(figure):
    kernels = [remove_reducedim, remove_reducedim_boost, remove_pd, remove_looped, remove_looped_boost, remove_smalldata]
    return {'setup': setup,
    'kernels': {'A': kernels, 'B': kernels[:3]}[figure],
    'n_range': {'A': [2 ** k for k in range(12)], 'B': [2 ** k for k in range(11, 25)]}[figure],
     'xlabel': 'Remowing n rows from n rows',
     'title' : {'A':'Testing removal of small dataset', 'B':'Testing removal of large dataset'}[figure],
     'show_progress': False,
     'equality_check': lambda x,y: np.array_equal(x, y)}
    
t = time()
outs = [perfplot.bench(**build_args(n)) for n in ('A','B')]
fig = plt.figure(figsize=(20, 20))
for i in range(len(outs)):
    ax = fig.add_subplot(2, 1, i+1)
    ax.grid(True, which="both")
    outs[i].plot()
plt.show()
print('Overall testing time:', time()-t)

输出：

Overall testing time: 529.2596168518066

Answer 1

这是使用and进行“反连接”的熊猫方法。mergequery

dfa = pd.DataFrame(a)
dfb = pd.DataFrame(b)

df = (
    dfa.merge(dfb, how='left', indicator=True)
    .query('_merge == "left_only"')
    .drop(columns='_merge')
)

    0   1   2   3
1  75   7   0  85
4  75  48   8  43
5   7  47  96  94
6   7  47  96  94
8  19  49  92  57

注意：普通的 numpy 解决方案应该更快，但这应该没问题。

普通的numpy但只有一个循环：

for arr in b:
    a = a[np.all(~np.equal(a, arr), axis=1)]

array([[75,  7,  0, 85],
       [75, 48,  8, 43],
       [ 7, 47, 96, 94],
       [ 7, 47, 96, 94],
       [19, 49, 92, 57]])

Answer 2

您可以通过两个矩阵中的向量之间的循环比较来尝试我的解决方案。

代码

import np

a = np.array([[25, 83, 18, 71],
       [75,  7,  0, 85],
       [25, 83, 18, 71],
       [25, 83, 18, 71],
       [75, 48,  8, 43],
       [ 7, 47, 96, 94],
       [ 7, 47, 96, 94],
       [56, 75, 50,  0],
       [19, 49, 92, 57],
       [52, 93, 58,  9]])

b = np.array([[56, 75, 50,  0], [52, 93, 58,  9], [25, 83, 18, 71]])

to_remain = [True]*len(a)

ind = 0

for vec_a in a:
  for vec_b in b:
    if np.array_equal(vec_a, vec_b):
      to_remain[ind] = False
      break
  
  ind += 1

output = a[to_remain]

print(output)

输出

[[75  7  0 85]
 [75 48  8 43]
 [ 7 47 96 94]
 [ 7 47 96 94]
 [19 49 92 57]]