将元组与 numpy 数组中的元组进行比较 [英] compare tuple with tuples in numpy array

问题描述

我有一个数组 (dtype=object)，第一列包含数组元组，第二列包含标量.我想要第二列中的所有标量，其中第一列中的元组等于某个元组.

说

<预><代码>>>>X数组([[(数组([ 21.]), 数组([ 13.])), 0.29452519286647716],[(数组([ 25.]), 数组([ 9.])), 0.9106600600510809],[(数组([ 25.]), 数组([ 13.])), 0.8137344043493814],[(数组([ 25.]), 数组([ 14.])), 0.8143093864975313],[(数组([ 25.]), 数组([ 15.])), 0.6004337591112664],[(数组([ 25.]), 数组([ 16.])), 0.6239450452872853],[(数组([ 21.]), 数组([ 13.])), 0.32082105959687424]], dtype=object)

我想要第一列等于 X[0,0] 的所有行.

ar = X[0,0]>>>ar(数组([ 21.]), 数组([ 13.]))

我虽然检查 X[:,0]==ar 应该能找到那些行.然后我会通过 X[X[:,0]==ar,1] 检索我的最终结果.

然而，似乎发生的是 ar 被解释为一个二维数组，并且 ar 中的每个元素都与 中的元组进行比较X[:,0].在这种情况下，这会产生一个 2x7 数组，所有条目都等于 False.相比之下，比较 X[0,0]==ar 就像我希望它给出 True 的值一样工作.

为什么会发生这种情况，我该如何修复它以获得所需的结果?

解决方案

比较使用列表推导式工作:

In [176]: [x==ar for x in X[:,0]]出[176]:[真，假，假，假，假，假，真]

这是元组与元组的比较

比较元组 ID 会得到不同的结果

In [175]: [id(x)==id(ar) for x in X[:,0]]出[175]:[真，假，假，假，假，假，假]

因为第二场比赛的 ID 不同.

在 [177]: X[:,0]==ar出[177]:数组([[假，假，假，假，假，假，假]，[假，假，假，假，假，假，假]]，dtype=bool)

返回一个 (2,7) 结果，因为它是将 (7,) 数组与 (2,1) 数组进行比较的效果> 数组 (np.array(ar)).

但这就像理解一样:

在[190]中:ar1=np.zeros(1,dtype=object)在 [191] 中:ar1[0]=ar在 [192]: ar1出[192]:数组([(数组([21.])，数组([13.]))]，dtype=object)在 [193] 中:X[:,0]==ar1出[193]:数组([真，假，假，假，假，假，真]，dtype=bool)

art1 是一个包含 ar 元组的 1 元素数组.现在与 X[:,0] 元素的比较按预期进行.

np.array(...) 尝试创建一个输入数据允许的尽可能高的维度数组.这就是为什么它将 2 元素元组转换为 2 元素数组.我不得不做一个两步分配来绕过这个默认设置.

I have an array (dtype=object) with the first column containing tuples of arrays and the second column containing scalars. I want all scalars from the second column where the tuples in the first column equal a certain tuple.

Say

>>> X
array([[(array([ 21.]), array([ 13.])), 0.29452519286647716],
   [(array([ 25.]), array([ 9.])), 0.9106600600510809],
   [(array([ 25.]), array([ 13.])), 0.8137344043493814],
   [(array([ 25.]), array([ 14.])), 0.8143093864975313],
   [(array([ 25.]), array([ 15.])), 0.6004337591112664],
   [(array([ 25.]), array([ 16.])), 0.6239450452872853],
   [(array([ 21.]), array([ 13.])), 0.32082105959687424]], dtype=object)

and I want all rows where the 1st column equals X[0,0].

ar = X[0,0]
>>> ar
(array([ 21.]), array([ 13.]))

I thaugh checking X[:,0]==ar should find me those rows. I would had then retrieved my final result by X[X[:,0]==ar,1].

What seems to happen, however, is that ar gets to be interpreted as a 2dimensional array and each single element in ar is compared to the tuples in X[:,0]. This yields a, in this case, 2x7 array all entries equal to False. In contrast, the comparison X[0,0]==ar works just as I would want it giving a value of True.

Why is that happening and how can I fix it to obtain the desired result?

解决方案

Comparison using list comprehension works:

In [176]: [x==ar for x in X[:,0]]
Out[176]: [True, False, False, False, False, False, True]

This is comparing tuples with tuples

Comparing tuple ids gives a different result

In [175]: [id(x)==id(ar) for x in X[:,0]]
Out[175]: [True, False, False, False, False, False, False]

since the 2nd match has a different id.

In [177]: X[:,0]==ar
Out[177]: 
array([[False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False]], dtype=bool)

returns a (2,7) result because it is, effect comparing a (7,) array with a (2,1) array (np.array(ar)).

But this works like the comprehension:

In [190]: ar1=np.zeros(1,dtype=object)

In [191]: ar1[0]=ar

In [192]: ar1
Out[192]: array([(array([ 21.]), array([ 13.]))], dtype=object)

In [193]: X[:,0]==ar1
Out[193]: array([ True, False, False, False, False, False,  True], dtype=bool)

art1 is a 1 element array containing the ar tuple. Now the comparison with the elements of X[:,0] proceeds as expected.

np.array(...) tries to create as high a dimension array as the input data allows. That is why it turns a 2 element tuple into a 2 element array. I had to do a 2 step assignment to get around that default.

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