"less"中的无效值在数组中比较np.nan时 [英] invalid value in "less" when comparing np.nan in an array

问题描述

观察

np.nan < 0

False

---

np.array([np.nan]) < 0

array([False], dtype=bool)

---

但是

---

However

np.array([-1, np.nan]) < 0

//anaconda/envs/3.5/lib/python3.5/site-packages/ipykernel/__main__.py:1: RuntimeWarning: invalid value encountered in less
  if \__name\__ == '\__main\__':

array([ True, False], dtype=bool)

---

该错误是什么意思?

---

What does that error mean?

推荐答案

这仅是警告，并非例外. Numpy会尝试提供帮助，并报告您数组中有一个nan.忽略它是安全的.如果警告本身困扰您(或您的用户)，则可以取消警告:

This is only a warning, not an exception. Numpy tries to be helpful and reports that you have a nan in your array. It's safe to ignore it. If the warning itself bothers you (or your users), you can suppress it:

with np.errstate(invalid='ignore'):
    np.less([np.nan, 0], 1)

但是，实际上不建议这样做，因为如果需要在冗长的计算中查找nan的来源，它可以掩盖不太明显的问题并使调试变得更加困难.

However, this is really not recommended since it can mask less obvious issues and make debugging harder if you need to find where a nan comes from in a lengthy calculation.

请注意nan的语义:nan > 0，nan < 0和nan < nan，nan == nan都是False.

Note the semantics of a nan: nan > 0, nan < 0 and nan < nan, nan == nan are all False.

还请注意，在numpy中(不同于熊猫)，nan的确表示无效而不是不可用.

Also note that in numpy (unlike e.g. pandas) nan really means invalid not not available.

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