问题描述
我有一些方法可以将不同数量的混乱数据传递给此函数,以将标头与数据组合并返回字典列表:
I have a few methods which pass different amount of messy data to this function to combine headers with data and return a list of dictionaries:
def zip_data(self, indicator_names, indicator_values):
values = [[float(elem) for elem in item] for item in np.nditer(indicator_values)]
return [dict(zip(indicator_names, row)) for row in values]
基本上就像(他们做的一样):
It's basically something like (they do the same):
def zip_large_data(self, indicator_names, indicator_values):
data = []
for item in np.nditer(indicator_values):
values = []
values.append(int(item[0]))
for elem in item[1:]:
values.append(float(elem))
data.append(dict(zip(indicator_names, values)))
return data
问题是,如果传递了20个元素的列表,则效果很好,但是对于40个元素,它会给出错误消息:
The thing is, it works great if a list of 20 elements is passed, but for like 40 it gives the error:
文件"xy.py",第51行,位于zip_large_data中 对于np.nditer(indicator_values)中的项目:
File "xy.py", line 51, in zip_large_data for item in np.nditer(indicator_values):
ValueError:操作数太多
ValueError: Too many operands
np.nditer()可以迭代多少个值?有什么办法可以避免这种情况?
How many values can np.nditer() iterate over? Is there any way to avoid this?
小例子:
indicator_names = ['a','b']
indicator_names = ['a', 'b']
想要的输出:
当前状态:
def zip_large_data(self, indicator_names, indicator_values):
data = []
print(indicator_values[0])
for item in np.nditer(indicator_values):
print(item)
values = []
values.append(int(item[0]))
for elem in item[1:]:
values.append(float(elem))
data.append(dict(zip(indicator_names, values)))
print(data)
break
return data
输出:
出局:[1 2 3 4 5]
Out: [1 2 3 4 5]
在:打印(项目)
输出:(array(1),array(5))
Out:(array(1), array(5))
在:打印(数据)
出局:[{'a':1,'b':5}]
Out: [{'a': 1, 'b': 5}]
所以基本上我不想顺序遍历indicator_values,而是每个数组的第一个元素,然后每个数组的第二个元素,等等.我想避免nditer,但不知道怎么做
So basically I do not want to iterate through the indicator_values sequentially, but first elements of every array, then the second elements of every array etc.. I want to avoid nditer, but don't see how
抱歉,英语不是我的母语,第一次使用numpy令人困惑.
Sorry English is not my first language, first time working with numpy, it's confusing.
推荐答案
您正在达到NPY_MAXARGS
限制.
我还没有看到nditer
这样的用法,所以花了我一些时间才弄清楚发生了什么.然后,我使用python会话测试了我的想法.一个可行的例子会有所帮助.
I haven't seen a use like this for nditer
, so it took me a bit to figure out what was happening. And then I had use a python session to test my ideas. A worked example would have helped.
通常,发帖人使用nditer
作为简单地遍历数组并进行一些计算的方式.简单迭代(无需nditer) is usually faster.
nditer is mainly a stepping stone toward implementing the idea in
cython`.
Usually posters use nditer
as a way of simply iterating through an array and do some calculation. Simple iteration (without nditer) is usually faster.
nditeris mainly a stepping stone toward implementing the idea in
cython`.
使用数组列表,nditer
一起广播它们,然后遍历匹配的元素.类似于常见的Python列表zip习惯用法(函数名暗含).
With a list of arrays, nditer
broadcasts them together, and then iterates through the matching elements. It's akin to the common Python list zip idiom (as implied by your function name).
In [152]: list(zip('abc',[1,2,3]))
Out[152]: [('a', 1), ('b', 2), ('c', 3)]
In [153]: {k:v for k,v in zip('abc',[1,2,3])}
Out[153]: {'a': 1, 'b': 2, 'c': 3}
定义3个可以相互广播的小阵列:
Defining 3 small arrays that can be broadcast against each other:
In [136]: a = np.array([[1,2],[3,4]])
In [137]: b = np.array([[4],[5]])
In [138]: c = np.array([10])
In [140]: np.broadcast_arrays(a,b,c)
Out[140]:
[array([[1, 2],
[3, 4]]), array([[4, 4],
[5, 5]]), array([[10, 10],
[10, 10]])]
使用nditer
:
In [143]: for x in np.nditer([a,b,c]):
...: print(x)
...:
(array(1), array(4), array(10))
(array(2), array(4), array(10))
(array(3), array(5), array(10))
(array(4), array(5), array(10))
并带有您的功能:
In [155]: zip_large_data('abc',[a,b,c])
Out[155]:
[{'a': 1, 'b': 4.0, 'c': 10.0},
{'a': 2, 'b': 4.0, 'c': 10.0},
{'a': 3, 'b': 5.0, 'c': 10.0},
{'a': 4, 'b': 5.0, 'c': 10.0}]
如果运行ok,但我用32个操作数执行相同的迭代,但失败,则返回33
If I do same sort of iteration with 32 operands if runs ok, but fails with 33
In [160]: for x in np.nditer([a,b,c]*11):
...: pass
ValueError: Too many operands
numpy
的操作数限制为32(尺寸限制为32).它没有很好的文档记录,并且不会经常出现.我只在使用np.choose
的问题中看到过它.
numpy
has a 32 operand limit (and 32 dimension limit). It isn't well documented, and doesn't come up often. I've only seen it in questions using np.choose
.
替代numpy.choose,允许任意或至少32个以上的参数?
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