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This article introduces the relevant knowledge of "detailed explanation of Numpy library in Python". In the operation of actual cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

First of all, I need to understand what Numpy is. From my impression, it is generally mentioned that this tool is associated with machine learning. Of course, there is a lot of gap between the actual understanding and the actual understanding.

To put it simply, Python itself is not designed as a scientific computing language, but with the popularity of Python and a large number of expansion, the demand for this area will be increasing, so there is Numpy as a supplement.

You can use pip to download Numpy, and pip install numpy can be done with one command, about more than 10 M

The environment of Numpy is much easier to build than TensorFlow, and the way of verification is very simple, import is fine.

> import numpy

If you want to view the version, you can use the following ways.

> numpy.version.full_version

'1.13.3'

> > >

Of course, we can specify aliases, which is easier to use, as we have done in subsequent tests.

> import numpy as np

> np.version.full_version

'1.13.3'

Let's warm up first.

Get a string of integers in the range of 10, an array to be exact.

> a = np.arange (10)

> print a

[0 1 2 3 4 5 6 7 8 9]

Print it out and see it's an array.

> > type (a)

Then we change the pattern and transform the array into two parts, each of which is five.

> a = a.reshape (2BI 5)

> print a

[[0 1 2 3 4]

[5 6 7 8 9]]

You can continue to ReFactor, recompose 20 elements, and then divide them into four parts, each with five elements.

> a = np.arange (20)

> a = a.reshape (4pm 5)

> print a

[[0 1 2 3 4]

[5 6 7 8 9]

[10 11 12 13 14]

[15 16 17 18 19]]

You can also construct high-dimensional arrays, such as the following.

> a = a.reshape (2pm 2pm 5)

> print a

[0 1 2 3 4]

[5 6 7 8 9]]

[[10 11 12 13 14]

[15 16 17 18 19]

> > >

Not if the input (2, 2, 2, 6) is not allowed, the obvious 2, 2, 2, 6, 24, has overflowed.

> a = a.reshape (2pm 2pm 6)

Traceback (most recent call last):

File "", line 1, in

ValueError: cannot reshape array of size 20 into shape (2mai 2pm 6)

> > >

We get the dimensions of the array

> a.ndim

three

Get the size of each dimension of the array

> a.shape

(2L, 2L, 5L)

View the number of elements in the array

> a.size

twenty

Then we create an array that can convert the list to get

> < raw >

> a = np.array (raw)

> > a

Array ([0,1,2,3,4])

> print a

[0 1 2 3 4]

It's the same operation for a slightly more complicated one.

> raw = [0pr 1, 2, 3, 4], [5, 6, 7, 7, 8, 9]]

B = np.array (raw)

> > b

Array ([0,1,2,3,4]

[5, 6, 7, 8, 9])

> > >

If you want to get an array displayed as 0, you can use this way to zeros

> d = (4pc5)

> > np.zeros (d)

Array ([[0, 0, 0, 0, 0.]

[0., 0., 0., 0.]

[0., 0., 0., 0.]

[0.,0.,0.,0.,0.])

Of course, the effect is not ideal, which shows the decimal point, which can be formatted and changed to an integer.

> np.ones (dforce dtypewriter int)

Array ([1,1,1,1,1]

[1, 1, 1, 1, 1]

[1, 1, 1, 1, 1]

[1,1,1,1,1])

Get random numbers and generate five in one breath.

> > np.random.rand (5)

Array ([0.61643689, 0.15915655, 0.20558268, 0.75157157, 0.50395262])

According to this idea, randomly generating 100 is also seconds.

Let's take a look at array operations.

This part gradually shows its advantage and can support computing.

> a = np.array ([[1jue 2], [2je 5]])

> print a

[[1 2]

[2 5]]

> b = np.array ([[2jue 3], [5je 8]])

> print aquib

[[3 5]

[7 13]]

Get some maximum and minimum values.

> a = np.arange (20) .reshape (4jue 5)

> print a

[[0 1 2 3 4]

[5 6 7 8 9]

[10 11 12 13 14]

[15 16 17 18 19]]

> > print str (a.sum ())

one hundred and ninety

> > print str (a.max ())

nineteen

> > print str (a.min ())

0

> > print str (a.max (axis=1))

[4 9 14 19]

> > >

It can be easily changed.

> b = np.arange (2pr 45pr 3) .reshape (5pr 3)

> b = np.mat (b)

> > print b

[[2 5 8]

[11 14 17]

[20 23 26]

[29 32 35]

[38 41 44]]

If the maximum number is 2, starting with 0, taking five numbers is written like this

> np.linspace (0pje 2pm 5)

Array ([0. , 0.5, 1., 1.5, 2.])

If it is nine numbers, it is as follows.

> np.linspace (0pje 2je 9)

Array ([0. , 0.25, 0.5, 0.75, 1., 1.25, 1.5, 1.75, 2.])

> > >

> a = np.array ([[3jue 2], [5je 9]])

> print a [0] [1]

two

> print a [1] [0]

five

> print a [1J 0]

five

> b = a

> > a [0] [1] = 100

> print a

[[3 100]

[5 9]]

> > print b

[[3 100]

[5 9]]

The above result may seem strange, but the reason for the problem is that instead of actually copying a to b, Python points b to the memory address of a's corresponding data.

You can use copy to circumvent.

> a = np.array ([[3jue 2], [5je 9]])

> b = a.copy ()

> > a [0] [1] = 100

> print a

[[3 100]

[5 9]]

> > print b

[[3 2]

[5 9]]

Specify column

> a = np.arange (20) .reshape (4jue 5)

> print a

[[0 1 2 3 4]

[5 6 7 8 9]

[10 11 12 13 14]

[15 16 17 18 19]]

> print a [:, [1jol 3]]

[[1 3]

[6 8]

[11 13]

[16 18]]

Splice two vectors into a matrix by column:

> a = np.array ((1 ~ 2 ~ 3))

> b = np.array ((2d3))

> print np.column_stack ((aPerm b))

[[1 2]

[2 3]

[3 4]]

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