Python 如何使用索引和值迭代 1d NumPy 数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/49384682/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to iterate 1d NumPy array with index and value
提问by user40780
For python dict, I could use iteritems()to loop through key and value at the same time. But I cannot find such functionality for NumPy array. I have to manually track idxlike this:
对于 python dict，我可以使用iteritems()同时循环遍历键和值。但是我找不到 NumPy 数组的这种功能。我必须idx像这样手动跟踪：
idx = 0
for j in theta:
some_function(idx,j,theta)
idx += 1
Is there a better way to do this?
有一个更好的方法吗？
回答by jpp
There are a few alternatives. The below assumes you are iterating over a 1d NumPy array.
有几种选择。下面假设您正在迭代一维 NumPy 数组。
Iterate with range
迭代 range
for j in range(theta.shape[0]): # or range(len(theta))
some_function(j, theta[j], theta)
Note this is the onlyof the 3 solutions which will work with numba. This is noteworthy since iterating over a NumPy array explicitly is usually only efficient when combined with numbaor another means of pre-compilation.
请注意，这是3 种解决方案中唯一可以与numba. 这是值得注意的，因为显式迭代 NumPy 数组通常仅在与numba预编译或其他预编译方法结合使用时才有效。
Iterate with enumerate
迭代 enumerate
for idx, j in enumerate(theta):
some_function(idx, j, theta)
The most efficient of the 3 solutions for 1d arrays. See benchmarking below.
1d 阵列的 3 种解决方案中最有效的。请参阅下面的基准测试。
Iterate with np.ndenumerate
迭代 np.ndenumerate
for idx, j in np.ndenumerate(theta):
some_function(idx[0], j, theta)
Notice the additional indexing step in idx[0]. This is necessary since the index (like shape) of a 1d NumPy array is given as a singleton tuple. For a 1d array, np.ndenumerateis inefficient; its benefits only show for multi-dimensional arrays.
请注意 中的附加索引步骤idx[0]。这是必要的，因为shape1d NumPy 数组的索引（如）是作为单例元组给出的。对于一维数组，np.ndenumerate效率低下；它的好处只显示在多维数组中。
Performance benchmarking
性能基准测试
# Python 3.7, NumPy 1.14.3
np.random.seed(0)
arr = np.random.random(10**6)
def enumerater(arr):
for index, value in enumerate(arr):
index, value
pass
def ranger(arr):
for index in range(len(arr)):
index, arr[index]
pass
def ndenumerater(arr):
for index, value in np.ndenumerate(arr):
index[0], value
pass
%timeit enumerater(arr) # 131 ms
%timeit ranger(arr) # 171 ms
%timeit ndenumerater(arr) # 579 ms
回答by sushmit
You can use numpy.ndenumeratefor example
你可以使用numpy.ndenumerate例如
import numpy as np
test_array = np.arange(2, 3, 0.1)
for index, value in np.ndenumerate(test_array):
print(index[0], value)
For more information refer to https://docs.scipy.org/doc/numpy/reference/generated/numpy.ndenumerate.html
有关更多信息，请参阅https://docs.scipy.org/doc/numpy/reference/generated/numpy.ndenumerate.html
