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This article will explain in detail how to operate in from Python to NumPy, which is closest to human thinking. The content of the article is of high quality, so the editor will share it for you to do a reference. I hope you will have a certain understanding of the relevant knowledge after reading this article.

In the Python language, in is a very frequently used operator to determine whether an object is in a string, tuple, list, collection, or dictionary. The operation of in is highly consistent with the human way of thinking, and its writing is close to natural language, which fully embodies the philosophy of Python.

> 'or' in' hello world' True > 5 in {1 name':'Mike', 2 age':18 3} False > 'age' in {' name':'Mike', 'age':18} True

Interestingly, with the exception of R, javascript, and SQL, there is little support for in operations in major languages, including C _ +. This may explain why the Python language is considered the easiest programming language to learn.

Accustomed to using Python's in operator, sometimes it is naturally applied to NumPy array operations. For example, there seems to be no problem with the following writing.

> import numpy as np > a = np.arange (9) > an array ([0,1,2,3,4,5,6,7,8]) > 5 in a True > 10 in a False

However, an accident occurred when I tried to use the in operator in the np.where () function.

> np.where (a > 5) (array ([6, 7, 8], dtype=int64),) > np.where (a% 2% 5) (array ([0, 2, 4, 6, 8], dtype=int64),) > > np.where (an in [2, 4, 6, 8]) Traceback (most recent call last): File ", line 1, in np.where (an in [2, 3, 6, 8]) ValueError: The truth value of an array with more than one element is ambiguous. Use a.any () or a.all ()

There is no problem with the np.where () function if a > 5 or a% 2 percent zero is used as a condition, but np.where () throws an exception when using an in [2jing3jin5jin7]. Even if it is written as follows, you can't get the desired results.

> np.where (an in np.array ([2min3, 5) (array ([], dtype=int64),)

Doesn't NumPy support in operations between two arrays? No, NumPy, which is so powerful that the universe is invincible, how can it not support in operation between arrays? NumPy not only supports it, but also supports it very well.

> p = np.array # prime array > np.in1d (a, p) # returns whether each element of an is a prime Boolean array array ([False, False, True, True, False, True, False, True, False]) > > np.where (np.in1d (a, p)) # returns the index ordinal number of prime numbers in the array a (array ([2meme 3meme 5Power7], dtype=int64) ) > np.where (np.in1d (a, p),-1, a) # returns the result array ([0, 1,-1,-1, 4,-1, 6,-1, 8]) where all prime numbers in array an are replaced by-1

If the argument of np.in1d () is a multi-dimensional array, it is automatically flattened, and the returned Boolean array is also a flattened one-dimensional array.

> np.in1d (a.reshape ((3Power3)), p) array ([False, False, True, True, False, True, False, True, False])

If the arguments to np.in1d () are multidimensional and you expect to return a Boolean array with the same structure as the original array, you should use the np.isin () function.

> np.isin (a.reshape ((3Power3)), p) array ([[False, False, True], [True, False, True], [False, True, False]]) > np.where (np.isin (a.reshape (3Power3)), p)) (array ([0,1,2], dtype=int64), array ([2,0,2,1], dtype=int64))

If you expect to get the intersection of two arrays instead of the index of the intersection elements, the following two ways are feasible.

> a [np.where (np.isin (a, p)] array ([2,3,5,7]) > np.intersect1d (a, p) array ([2,3,5,7])

The second way is to directly use the np.intersect1d () function to get the intersection of the two arrays and sort them automatically. But I prefer the first way.

About from Python to NumPy closest to the human mind in operation is shared here, I hope that the above content can be of some help to you, can learn more knowledge. If you think the article is good, you can share it for more people to see.

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