作为这个问题的后续(感谢MSeifert的帮助(，我遇到了一个问题，即在传递掩码数组以更新val_dict之前，必须用索引数组new_vals_idx掩码numpy数组new_values。
对于MSeifert在旧帖子中提出的解决方案，我试图应用阵列掩蔽，但性能并不令人满意
我在以下示例中使用的数组和dicts是：
import numpy as np
val_dict = {'a': 5.0, 'b': 18.8, 'c': -55/2}
for i in range(200):
val_dict[str(i)] = i
val_dict[i] = i**2
keys = ('b', 123, '89', 'c') # dict keys to update
new_values = np.arange(1, 51, 1) / 1.0 # array with new values which has to be masked
new_vals_idx = np.array((0, 3, 5, -1)) # masking array
valarr = np.zeros((new_vals_idx.shape[0])) # preallocation for masked array
length = new_vals_idx.shape[0]
为了使我的代码片段更容易与以前的问题进行比较，我将坚持使用MSeifert答案的函数命名。以下是我试图从python/cython中获得最佳性能的尝试(其他答案由于性能太差而被忽略(：
def old_for(val_dict, keys, new_values, new_vals_idx, length):
for i in range(length):
val_dict[keys[i]] = new_values[new_vals_idx[i]]
%timeit old_for(val_dict, keys, new_values, new_vals_idx, length)
# 1000000 loops, best of 3: 1.6 µs per loop
def old_for_w_valarr(val_dict, keys, new_values, valarr, new_vals_idx, length):
valarr = new_values[new_vals_idx]
for i in range(length):
val_dict[keys[i]] = valarr[i]
%timeit old_for_w_valarr(val_dict, keys, new_values, valarr, new_vals_idx, length)
# 100000 loops, best of 3: 2.33 µs per loop
def new2_w_valarr(val_dict, keys, new_values, valarr, new_vals_idx, length):
valarr = new_values[new_vals_idx].tolist()
for key, val in zip(keys, valarr):
val_dict[key] = val
%timeit new2_w_valarr(val_dict, keys, new_values, valarr, new_vals_idx, length)
# 100000 loops, best of 3: 2.01 µs per loop
Cython函数：
%load_ext cython
%%cython
import numpy as np
cimport numpy as np
cpdef new3_cy(dict val_dict, tuple keys, double[:] new_values, int[:] new_vals_idx, Py_ssize_t length):
cdef Py_ssize_t i
cdef double val # this gives about 10 µs speed boost compared to directly assigning it to val_dict
for i in range(length):
val = new_values[new_vals_idx[i]]
val_dict[keys[i]] = val
%timeit new3_cy(val_dict, keys, new_values, new_vals_idx, length)
# 1000000 loops, best of 3: 1.38 µs per loop
cpdef new3_cy_mview(dict val_dict, tuple keys, double[:] new_values, int[:] new_vals_idx, Py_ssize_t length):
cdef Py_ssize_t i
cdef int[:] mview_idx = new_vals_idx
cdef double [:] mview_vals = new_values
for i in range(length):
val_dict[keys[i]] = mview_vals[mview_idx[i]]
%timeit new3_cy_mview(val_dict, keys, new_values, new_vals_idx, length)
# 1000000 loops, best of 3: 1.38 µs per loop
# NOT WORKING:
cpdef new2_cy_mview(dict val_dict, tuple keys, double[:] new_values, int[:] new_vals_idx, Py_ssize_t length):
cdef double [new_vals_idx] masked_vals = new_values
for key, val in zip(keys, masked_vals.tolist()):
val_dict[key] = val
cpdef new2_cy_mask(dict val_dict, tuple keys, double[:] new_values, valarr, int[:] new_vals_idx, Py_ssize_t length):
valarr = new_values[new_vals_idx]
for key, val in zip(keys, valarr.tolist()):
val_dict[key] = val
Cython函数new3_cy和new3_cy_mview似乎并不比old_for快得多。传递valarr以避免函数内部的数组构造(因为它将被调用数百万次(甚至似乎会减慢它的速度
在new2_cy_mask中使用Cython中的new_vals_idx数组进行掩码时会出现错误："指定的内存视图的索引无效，请键入int[：]"。对于索引数组，有没有类似Py_ssize_t的类型
尝试在new2_cy_mview中创建屏蔽内存视图时，会出现错误"无法将类型'double[：]'分配给'double[__pyx_v_new_vals_idx]'"。有没有类似于蒙面记忆视图的东西？我找不到关于这个主题的信息。。。
将计时结果与我以前的问题中的结果进行比较，我想数组屏蔽是占用大部分时间的过程。由于它很可能已经在numpy中进行了高度优化，所以可能没有什么可做的。但速度太慢了，必须(希望(有更好的方法来做到这一点。
任何帮助都将不胜感激！提前感谢！
在当前构造中可以做的一件事是关闭边界检查(如果安全的话！(。不会有太大的不同，但会有一些增量性能。
%%cython
import numpy as np
cimport numpy as np
cimport cython
@cython.boundscheck(False)
@cython.wraparound(False)
cpdef new4_cy(dict val_dict, tuple keys, double[:] new_values, int[:] new_vals_idx, Py_ssize_t length):
cdef Py_ssize_t i
cdef double val # this gives about 10 µs speed boost compared to directly assigning it to val_dict
for i in range(length):
val = new_values[new_vals_idx[i]]
val_dict[keys[i]] = val
In [36]: %timeit new3_cy(val_dict, keys, new_values, new_vals_idx, length)
1.76 µs ± 209 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [37]: %timeit new4_cy(val_dict, keys, new_values, new_vals_idx, length)
1.45 µs ± 31.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)