Chapter one Preliminary knowledge

## One 、Python Basics ### 1. List derivation and conditional assignment

When generating a sequence of numbers , stay Python It can be written as follows ：

L = []
def my_func(x):
    return 2*x
for i in range(5):
    L.append(my_func(i))
L
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[0, 2, 4, 6, 8]
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In fact, we can use the list derivation to simplify the writing method ：[* for i in *]. among , first * For mapping function , Its input is the following i Content of reference , the second * Object representing iteration .

[my_func(i) for i in range(5)]
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[0, 2, 4, 6, 8]
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List expressions also support multi-level nesting , As the first in the following example for For outer circulation , The second is the inner loop ：

[m+'_'+n for m in ['a', 'b'] for n in ['c', 'd']]
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['a_c', 'a_d', 'b_c', 'b_d']
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Except for list derivation , Another useful grammar is sugar with if Assignment of selected conditions , In the form of value = a if condition else b：

value = 'cat' if 2>1 else 'dog'
value
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'cat'
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Equivalent to the following expression ：

a, b = 'cat', 'dog'
condition = 2 > 1 #  This is the case True
if condition:
    value = a
else:
    value = b
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Here's an example , More than... In the truncated list 5 The elements of , More than 5 The use of 5 Instead of , Less than 5 Keep the original value ：

L = [1, 2, 3, 4, 5, 6, 7]
[i if i <= 5 else 5 for i in L]
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[1, 2, 3, 4, 5, 5, 5]
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2. Anonymous function and map Method

Some functions are defined with clear and simple mapping relationships , For example, above my_func function , At this time, we can use the method of anonymous function to express ：

my_func = lambda x: 2*x
my_func(3)
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6
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multi_para_func = lambda a, b: a + b
multi_para_func(1, 2) 
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3
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But the above usage actually violates “ anonymous ” The meaning of , In fact, it is often used in situations where multiple calls are not required , For example, the example in the above list derivation , The user doesn't care about the name of the function , Only care about this mapping relationship ：

[(lambda x: 2*x)(i) for i in range(5)]
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[0, 2, 4, 6, 8]
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For the anonymous function mapping of the above list derivation ,Python Provided in map Function to complete , What it returns is a map object , Need to pass through list Turn to list ：

list(map(lambda x: 2*x, range(5)))
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[0, 2, 4, 6, 8]
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For function mapping of multiple input values , You can do this by appending an iteration object ：

list(map(lambda x, y: str(x)+'_'+y, range(5), list('abcde')))
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['0_a', '1_b', '2_c', '3_d', '4_e']
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3. zip Object and the enumerate Method

zip Function can package multiple iteratable objects into an iteratable object composed of a tuple , It returns a zip object , adopt tuple, list You can get the corresponding packaging results ：

L1, L2, L3 = list('abc'), list('def'), list('hij')
list(zip(L1, L2, L3))
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[('a', 'd', 'h'), ('b', 'e', 'i'), ('c', 'f', 'j')]
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tuple(zip(L1, L2, L3))
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(('a', 'd', 'h'), ('b', 'e', 'i'), ('c', 'f', 'j'))
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It is often used in loop iteration zip function ：

for i, j, k in zip(L1, L2, L3):
     print(i, j, k)
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a d h
b e i
c f j
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enumerate It's a special kind of packing , It can bind the traversal sequence number of iterative elements during iteration ：

L = list('abcd')
for index, value in enumerate(L):
     print(index, value)
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0 a
1 b
2 c
3 d
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use zip Objects can also simply implement this function ：

for index, value in zip(range(len(L)), L):
     print(index, value)
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0 a
1 b
2 c
3 d
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When you need to establish a dictionary mapping between two lists , You can use zip object ：

dict(zip(L1, L2))
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{'a': 'd', 'b': 'e', 'c': 'f'}
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Now that you have the compression function , that Python Also provided * Operators, and zip Joint use to decompress ：

zipped = list(zip(L1, L2, L3))
zipped
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[('a', 'd', 'h'), ('b', 'e', 'i'), ('c', 'f', 'j')]
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list(zip(*zipped)) #  The three tuples correspond to the original list 
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[('a', 'b', 'c'), ('d', 'e', 'f'), ('h', 'i', 'j')]
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Two 、Numpy Basics

1. np Array construction

The most common way is through array To construct the ：

import numpy as np
np.array([1,2,3])
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array([1, 2, 3])
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Here are some ways to generate special arrays ：

【a】 The arithmetic sequence ：np.linspace, np.arange

np.linspace(1,5,11) #  start 、 End （ contain ）、 Number of samples 
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array([1. , 1.4, 1.8, 2.2, 2.6, 3. , 3.4, 3.8, 4.2, 4.6, 5. ])
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np.arange(1,5,2) #  start 、 End （ It doesn't contain ）、 step 
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array([1, 3])
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【b】 Special matrix ：zeros, eye, full

np.zeros((2,3)) #  The incoming tuple represents the size of each dimension 
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array([[0., 0., 0.],
       [0., 0., 0.]])
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np.eye(3) # 3*3 The identity matrix of 
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array([[1., 0., 0.],
       [0., 1., 0.],
       [0., 0., 1.]])
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np.eye(3, k=1) #  Offset the main diagonal 1 A pseudo identity matrix of units 
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array([[0., 1., 0.],
       [0., 0., 1.],
       [0., 0., 0.]])
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np.full((2,3), 10) #  Tuple incoming size ,10 Indicates the fill value 
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array([[10, 10, 10],
       [10, 10, 10]])
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np.full((2,3), [1,2,3]) #  Fill in the same list for each line 
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array([[1, 2, 3],
       [1, 2, 3]])
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【c】 Random matrix ：np.random

The most commonly used random generating function is rand, randn, randint, choice, They represent 0-1 Evenly distributed random arrays 、 Standard normal random array 、 Random integer group and random list sampling ：

np.random.rand(3) #  Generative obedience 0-1 Three random numbers uniformly distributed 
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array([0.92340835, 0.20019461, 0.40755472])
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np.random.rand(3, 3) #  Note that this is not a tuple , Each dimension size is entered separately 
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array([[0.8012362 , 0.53154881, 0.05858554],
       [0.13103034, 0.18108091, 0.30253153],
       [0.00528884, 0.99402007, 0.36348797]])
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For obedience interval a To b The uniform distribution on can be generated as follows ：

a, b = 5, 15
(b - a) * np.random.rand(3) + a
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array([6.59370831, 8.03865138, 9.19172546])
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General , You can select existing library functions ：

np.random.uniform(5, 15, 3)
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array([11.26499636, 13.12311185,  6.00774156])
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randn Generated N(0,I) The standard normal distribution of ：

np.random.randn(3)
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array([ 1.87000209,  1.19885561, -0.58802943])
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np.random.randn(2, 2)
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array([[-1.3642839 , -0.31497567],
       [-1.9452492 , -3.17272882]])
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For compliance, the variance is $\sigma^2$ The mean for $\mu$ The univariate normal distribution of can be generated as follows ：

sigma, mu = 2.5, 3
mu + np.random.randn(3) * sigma
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array([1.56024917, 0.22829486, 7.3764211 ])
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alike , You can also choose to generate... From existing functions ：

np.random.normal(3, 2.5, 3)
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array([3.53517851, 5.3441269 , 3.51192744])
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randint You can specify the minimum and maximum values for generating random integers （ It doesn't contain ） And dimension size ：

low, high, size = 5, 15, (2,2) #  Generate 5 To 14 Random integer of 
np.random.randint(low, high, size)
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array([[ 5, 12],
       [14,  9]])
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choice From a given list , Extract results in a certain probability and way , Uniform sampling when probability is not specified , The default sampling method is put back sampling ：

my_list = ['a', 'b', 'c', 'd']
np.random.choice(my_list, 2, replace=False, p=[0.1, 0.7, 0.1 ,0.1])
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array(['b', 'a'], dtype='<U1')
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np.random.choice(my_list, (3,3))
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array([['c', 'b', 'd'],
       ['d', 'a', 'd'],
       ['a', 'c', 'd']], dtype='<U1')
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When the number of returned elements is the same as the original list , Sampling without putting it back is equivalent to using permutation function , That is to break up the original list ：

np.random.permutation(my_list)
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array(['c', 'a', 'd', 'b'], dtype='<U1')
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Last , It should be mentioned that random seeds , It can fix the output of random numbers ：

np.random.seed(0)
np.random.rand()
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0.5488135039273248
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np.random.seed(0)
np.random.rand()
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0.5488135039273248
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2. np Deformation and merging of arrays

【a】 Transposition ：T

np.zeros((2,3)).T
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array([[0., 0.],
       [0., 0.],
       [0., 0.]])
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【b】 Merge operation ：r_, c_

For two-dimensional arrays ,r_ and c_ Represents up-down merging and left-right merging respectively ：

np.r_[np.zeros((2,3)),np.zeros((2,3))]
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array([[0., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.]])
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np.c_[np.zeros((2,3)),np.zeros((2,3))]
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array([[0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.]])
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When merging one-dimensional arrays and two-dimensional arrays , Think of it as a column vector , It can only be used in the case of left and right length matching c_ operation ：

try:
     np.r_[np.array([0,0]),np.zeros((2,1))]
except Exception as e:
     Err_Msg = e
Err_Msg
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ValueError('all the input arrays must have same number of dimensions, but the array at index 0 has 1 dimension(s) and the array at index 1 has 2 dimension(s)')
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np.r_[np.array([0,0]),np.zeros(2)]
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array([0., 0., 0., 0.])
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np.c_[np.array([0,0]),np.zeros((2,3))]
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array([[0., 0., 0., 0.],
       [0., 0., 0., 0.]])
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【c】 Dimensional transformation ：reshape

reshape It can help users rearrange the original array according to the new dimension . There are two modes in use , Respectively C Patterns and F Pattern , Fill and read in row by row and column by column respectively .

target = np.arange(8).reshape(2,4)
target
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array([[0, 1, 2, 3],
       [4, 5, 6, 7]])
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target.reshape((4,2), order='C') #  Read and fill by line 
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array([[0, 1],
       [2, 3],
       [4, 5],
       [6, 7]])
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target.reshape((4,2), order='F') #  Read and fill by column 
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array([[0, 2],
       [4, 6],
       [1, 3],
       [5, 7]])
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Specially , Since the size of the called array is determined ,reshape One dimension is allowed to be vacant , Just fill in -1 that will do ：

target.reshape((4,-1))
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array([[0, 1],
       [2, 3],
       [4, 5],
       [6, 7]])
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Next n*1 The size of the array is converted to 1 The operation of dimension group is often used ：

target = np.ones((3,1))
target
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array([[1.],
       [1.],
       [1.]])
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target.reshape(-1)
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array([1., 1., 1.])
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3. np Slicing and indexing of arrays

The slice mode of array supports the use of slice Type of start:end:step section , You can also directly pass in the list to specify the index of a dimension for slicing ：

target = np.arange(9).reshape(3,3)
target
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array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
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target[:-1, [0,2]]
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array([[0, 2],
       [3, 5]])
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Besides , It can also be used np.ix_ Use Boolean indexes on the corresponding dimensions , But you can't use slice section ：

target[np.ix_([True, False, True], [True, False, True])]
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array([[0, 2],
       [6, 8]])
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target[np.ix_([1,2], [True, False, True])]
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array([[3, 5],
       [6, 8]])
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When the array dimension is 1 Dimension time , Boolean indexing can be done directly , No need np.ix_：

new = target.reshape(-1)
new[new%2==0]
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array([0, 2, 4, 6, 8])
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4. Common functions

For the sake of simplicity , It is assumed that the input arrays of the following functions are one-dimensional .

【a】where

where Is a conditional function , You can specify the fill value corresponding to the position where the condition is met and the position where the condition is not met ：

a = np.array([-1,1,-1,0])
np.where(a>0, a, 5) #  The corresponding position is True Fill when a The corresponding element , Otherwise fill 5
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array([5, 1, 5, 5])
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【b】nonzero, argmax, argmin

All three functions return indexes ,nonzero Returns the index of a nonzero number ,argmax, argmin Returns the index of the largest and the smallest number respectively ：

a = np.array([-2,-5,0,1,3,-1])
np.nonzero(a)
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(array([0, 1, 3, 4, 5], dtype=int64),)
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a.argmax()
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4
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a.argmin()
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1
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【c】any, all

any When the sequence is at least There is one. True Or non-zero elements True, Otherwise return to False

all Refers to when the sequence element All for True Or non-zero elements True, Otherwise return to False

a = np.array([0,1])
a.any()
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True
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 a.all()
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False
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【d】cumprod, cumsum, diff

cumprod, cumsum Represents the cumulative multiplication and accumulation functions, respectively , Returns an array of the same length ,diff Represents the difference from the previous element , Because the first element is a missing value , So by default , The return length is the original array minus 1

a = np.array([1,2,3])
a.cumprod()
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array([1, 2, 6], dtype=int32)
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a.cumsum()
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array([1, 3, 6], dtype=int32)
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np.diff(a)
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array([1, 1])
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【e】 Statistical function

Common statistical functions include max, min, mean, median, std, var, sum, quantile, Quantile calculation is a global method , So it can't pass array.quantile Method call for ：

target = np.arange(5)
target
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array([0, 1, 2, 3, 4])
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target.max()
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4
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np.quantile(target, 0.5) # 0.5 quantile 
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2.0
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But for arrays with missing values , They also return missing values , If you need to skip missing values , You have to use nan* Function of type , The above statistical functions have corresponding nan* function .

target = np.array([1, 2, np.nan])
target
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array([ 1.,  2., nan])
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target.max()
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nan
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np.nanmax(target)
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2.0
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np.nanquantile(target, 0.5)
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1.5
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For covariance and correlation coefficient, we can use cov, corrcoef The calculation is as follows ：

target1 = np.array([1,3,5,9])
target2 = np.array([1,5,3,-9])
np.cov(target1, target2)
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array([[ 11.66666667, -16.66666667],
       [-16.66666667,  38.66666667]])
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np.corrcoef(target1, target2)
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array([[ 1.        , -0.78470603],
       [-0.78470603,  1.        ]])
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Last , It needs to be explained Numpy Of statistical functions in an array axis Parameters , It can calculate the statistical characteristics of a certain dimension , When axis=0 When the result is a column of statistical indicators , When axis=1 When the result is a row of statistical indicators ：

target = np.arange(1,10).reshape(3,-1)
target
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array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])
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target.sum(0)
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array([12, 15, 18])
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target.sum(1)
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array([ 6, 15, 24])
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5. Broadcast mechanism

The broadcast mechanism is used to handle the operation between two arrays of different dimensions , Here we only discuss the array broadcast mechanism which is no more than two dimensions .

【a】 Scalar and array operations

When a scalar and an array operate , The scalar will automatically expand the size to the size of the array , Then perform element by element operation ：

res = 3 * np.ones((2,2)) + 1
res
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array([[4., 4.],
       [4., 4.]])
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res = 1 / res
res
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array([[0.25, 0.25],
       [0.25, 0.25]])
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【b】 Operations between two-dimensional arrays

When two array dimensions are exactly the same , Use the operation of the corresponding element , Otherwise, an error will be reported , Unless the dimension of one of the arrays is $m×1$ perhaps $1×n$, Then it will expand its $1$ The dimension of is the size of the dimension corresponding to another array . for example , $1×2$ Array and $3×2$ When an array is operated on element by element, the first array will be expanded to $3×2$, Assign value to corresponding value during expansion . however , It should be noted that , If the dimension of the first array is $1×3$, Then, because the size on the second dimension does not match and is not $1$, An error at this time .

res = np.ones((3,2))
res
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array([[1., 1.],
       [1., 1.],
       [1., 1.]])
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res * np.array([[2,3]]) #  The second array expands the first dimension to 3
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array([[2., 3.],
       [2., 3.],
       [2., 3.]])
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res * np.array([[2],[3],[4]]) #  The second array expands the second dimension to 2
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array([[2., 2.],
       [3., 3.],
       [4., 4.]])
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res * np.array([[2]]) #  Equivalent to two extensions , The two dimensions of the second array are expanded into 3 and 2
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array([[2., 2.],
       [2., 2.],
       [2., 2.]])
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【c】 Operation of one-dimensional array and two-dimensional array

When a one-dimensional array $A_k$ And two-dimensional array $B_{m,n}$ In operation , It is equivalent to treating a one-dimensional array as $A_{1,k}$ Two dimensional array of , The broadcast rules used are the same as 【b】 In the agreement , When $k!=n$ And $k,n$ Are not $1$ Times wrong .

np.ones(3) + np.ones((2,3))
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array([[2., 2., 2.],
       [2., 2., 2.]])
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np.ones(3) + np.ones((2,1))
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array([[2., 2., 2.],
       [2., 2., 2.]])
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np.ones(1) + np.ones((2,3))
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array([[2., 2., 2.],
       [2., 2., 2.]])
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6. Calculation of vectors and matrices

【a】 Vector inner product ：dot

$\rm \mathbf{a}\cdot\mathbf{b} = \sum_ia_ib_i$

a = np.array([1,2,3])
b = np.array([1,3,5])
a.dot(b)
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22
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【b】 Vector norm and matrix norm ：np.linalg.norm

In the calculation of matrix norm , most important of all ord Parameters , The optional values are as follows ：

ord	norm for matrices	norm for vectors
None	Frobenius norm	2-norm
'fro'	Frobenius norm	/
'nuc'	nuclear norm	/
inf	max(sum(abs(x), axis=1))	max(abs(x))
-inf	min(sum(abs(x), axis=1))	min(abs(x))
0	/	sum(x != 0)
1	max(sum(abs(x), axis=0))	as below
-1	min(sum(abs(x), axis=0))	as below
2	2-norm (largest sing. value)	as below
-2	smallest singular value	as below
other	/	sum(abs(x)**ord)**(1./ord)

matrix_target =  np.arange(4).reshape(-1,2)
matrix_target
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array([[0, 1],
       [2, 3]])
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np.linalg.norm(matrix_target, 'fro')
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3.7416573867739413
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np.linalg.norm(matrix_target, np.inf)
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5.0
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np.linalg.norm(matrix_target, 2)
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3.702459173643833
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vector_target =  np.arange(4)
vector_target
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array([0, 1, 2, 3])
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np.linalg.norm(vector_target, np.inf)
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3.0
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np.linalg.norm(vector_target, 2)
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3.7416573867739413
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np.linalg.norm(vector_target, 3)
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3.3019272488946263
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【c】 Matrix multiplication ：@

$\rm [\mathbf{A}_{m\times p}\mathbf{B}_{p\times n}]_{ij} = \sum_{k=1}^p\mathbf{A}_{ik}\mathbf{B}_{kj}$

a = np.arange(4).reshape(-1,2)
a
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array([[0, 1],
       [2, 3]])
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b = np.arange(-4,0).reshape(-1,2)
b
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array([[-4, -3],
       [-2, -1]])
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[email protected]
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array([[ -2,  -1],
       [-14,  -9]])
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3、 ... and 、 practice

Ex1： Write matrix multiplication using list derivation

The general matrix multiplication is based on the formula , It can be written by a triple loop , Please rewrite it in the form of list derivation .

M1 = np.random.rand(2,3)
M2 = np.random.rand(3,4)
res = np.empty((M1.shape[0],M2.shape[1]))
for i in range(M1.shape[0]):
    for j in range(M2.shape[1]):
        item = 0
        for k in range(M1.shape[1]):
            item += M1[i][k] * M2[k][j]
        res[i][j] = item
(np.abs(([email protected] - res) < 1e-15)).all() #  Eliminate numerical errors 
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True
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Ex2： Update the matrix

Let's set the matrix $A_{m×n}$ , Right now $A$ Each element in the is updated to generate a matrix $B$ , The update method is $B_{ij}=A_{ij}\sum_{k=1}^n\frac{1}{A_{ik}}$ , For example, the following matrix is $A$ , be $B_{2,2}=5\times(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})=\frac{37}{12}$ , Please use Numpy Efficient implementation . \begin{split}A=\left[ \begin{matrix} 1 & 2 &3\\4&5&6\\7&8&9 \end{matrix} \right]\end{split}

Ex3： Chi square statistics

Let's set the matrix $A_{m\times n}$, remember $B_{ij} = \frac{(\sum_{i=1}^mA_{ij})\times (\sum_{j=1}^nA_{ij})}{\sum_{i=1}^m\sum_{j=1}^nA_{ij}}$, Define chi square values as follows ： $\chi^2 = \sum_{i=1}^m\sum_{j=1}^n\frac{(A_{ij}-B_{ij})^2}{B_{ij}}$ Please use Numpy For a given matrix $A$ Calculation $\chi^2$

np.random.seed(0)
A = np.random.randint(10, 20, (8, 5))
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Ex4： Improve the performance of matrix calculation

set up $Z$ by $m×n$ Matrix , $B$ and $U$ Namely $m×p$ and $p×n$ Matrix , $B_i$ by $B$ Of the $i$ That's ok , $U_j$ by $U$ Of the $j$ Column , The following definition $\displaystyle R=\sum_{i=1}^m\sum_{j=1}^n\|B_i-U_j\|_2^2Z_{ij}$, among $\|\mathbf{a}\|_2^2$ It's a vector $a$ Sum of squares of components $\sum_i a_i^2$.

Someone now calculates... Based on the sample data given below $R$ Value , Please make full use of Numpy The function in , Based on this problem, improve the performance of this code .

np.random.seed(0)
m, n, p = 100, 80, 50
B = np.random.randint(0, 2, (m, p))
U = np.random.randint(0, 2, (p, n))
Z = np.random.randint(0, 2, (m, n))
def solution(B=B, U=U, Z=Z):
    L_res = []
    for i in range(m):
        for j in range(n):
            norm_value = ((B[i]-U[:,j])**2).sum()
            L_res.append(norm_value*Z[i][j])
    return sum(L_res)
solution(B, U, Z)
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100566
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Ex5： The maximum length of a continuous integer

Enter an integer of Numpy Array , Returns the maximum length of a subarray of strictly incremented consecutive integers , Positive refers to the increasing direction . for example , Input [1,2,5,6,7],[5,6,7] Is a subarray of consecutive integers with maximum length , So output 3; Input [3,2,1,2,3,4,6],[1,2,3,4] Is a subarray of consecutive integers with maximum length , So output 4. Please make full use of Numpy The built-in function of .（ Tips ： Consider using nonzero, diff function ）