Question

我有一个大小（4，X，Y）的numpy数组，其中第一个维度代表（R，G，B，A）四元组。我的目的是将每个X*Y RGBA四元组转换为X*Y浮点值，给定与它们匹配的字典。

我目前的代码如下：

codeTable = {
    (255, 255, 255, 127): 5.5,
    (128, 128, 128, 255): 6.5,
    (0  , 0  , 0  , 0  ): 7.5,
}

for i in range(0, rows):
    for j in range(0, cols):
        new_data[i,j] = codeTable.get(tuple(data[:,i,j]), -9999)

其中data是一个大小为(4, rows, cols)的numpy数组，而new_data的大小为(rows, cols)。

代码工作正常，但需要很长时间。我该如何优化这段代码？

以下是一个完整的例子：

import numpy

codeTable = {
    (253, 254, 255, 127): 5.5,
    (128, 129, 130, 255): 6.5,
    (0  , 0  , 0  , 0  ): 7.5,
}

# test data
rows = 2
cols = 2
data = numpy.array([
    [[253, 0], [128,   0], [128,  0]],
    [[254, 0], [129, 144], [129,  0]],
    [[255, 0], [130, 243], [130,  5]],
    [[127, 0], [255, 120], [255,  5]],
])

new_data = numpy.zeros((rows,cols), numpy.float32)

for i in range(0, rows):
    for j in range(0, cols):
        new_data[i,j] = codeTable.get(tuple(data[:,i,j]), -9999)

# expected result for `new_data`:
# array([[  5.50000000e+00,   7.50000000e+00],
#        [  6.50000000e+00,  -9.99900000e+03],
#        [  6.50000000e+00,  -9.99900000e+03], dtype=float32)

Answer 1

这是一种返回预期结果的方法，但是如果数据量很少，很难知道这对您来说会更快。因为我已经避免了双循环，但我想你会看到相当不错的加速。

import numpy
import pandas as pd


codeTable = {
    (253, 254, 255, 127): 5.5,
    (128, 129, 130, 255): 6.5,
    (0  , 0  , 0  , 0  ): 7.5,
}

# test data
rows = 3
cols = 2
data = numpy.array([
    [[253, 0], [128,   0], [128,  0]],
    [[254, 0], [129, 144], [129,  0]],
    [[255, 0], [130, 243], [130,  5]],
    [[127, 0], [255, 120], [255,  5]],
])

new_data = numpy.zeros((rows,cols), numpy.float32)

for i in range(0, rows):
    for j in range(0, cols):
        new_data[i,j] = codeTable.get(tuple(data[:,i,j]), -9999)

def create_output(data):
    # Reshape your two data sources to be a bit more sane
    reshaped_data = data.reshape((4, -1))
    df = pd.DataFrame(reshaped_data).T

    reshaped_codeTable = []
    for key in codeTable.keys():
        reshaped = list(key) + [codeTable[key]]
        reshaped_codeTable.append(reshaped)
    ct = pd.DataFrame(reshaped_codeTable)

    # Merge on the data, replace missing merges with -9999
    result = df.merge(ct, how='left')
    newest_data = result[4].fillna(-9999)

    # Reshape
    output = newest_data.reshape(rows, cols)
    return output

output = create_output(data)
print(output)
# array([[  5.50000000e+00,   7.50000000e+00],
#        [  6.50000000e+00,  -9.99900000e+03],
#        [  6.50000000e+00,  -9.99900000e+03])

print(numpy.array_equal(new_data, output))
# True

Answer 2

numpy_indexed包（免责声明：我是它的作者）包含一个vector.ndex的矢量化nd数组变体，它可以用来有效和简洁地解决你的问题：

import numpy_indexed as npi
map_keys = np.array(list(codeTable.keys()))
map_values = np.array(list(codeTable.values()))
indices = npi.indices(map_keys, data.reshape(4, -1).T, missing='mask')
remapped = np.where(indices.mask, -9999, map_values[indices.data]).reshape(data.shape[1:])

提高numpy映射操作的性能

2 个答案: