np.vstack: Pinoamaan taulukoita pystyakselia pitkin. np.hstack: Pinoamaan taulukoita vaaka-akselia pitkin. np.column_stack: 1-D-taulukoiden pinoaminen sarakkeina 2-D-taulukoiksi. np.concatenate: Pinoamaan taulukoita määrättyä akselia pitkin (akseli välitetään argumenttina).

   import   numpy   as   np   a   =   np  .  array  ([[  1     2  ]   [  3     4  ]])   b   =   np  .  array  ([[  5     6  ]   [  7     8  ]])   # vertical stacking   print  (  'Vertical stacking:  n  '     np  .  vstack  ((  a     b  )))   # horizontal stacking   print  (  '  n  Horizontal stacking:  n  '     np  .  hstack  ((  a     b  )))   c   =   [  5     6  ]   # stacking columns   print  (  '  n  Column stacking:  n  '     np  .  column_stack  ((  a     c  )))   # concatenation method    print  (  '  n  Concatenating to 2nd axis:  n  '     np  .  concatenate  ((  a     b  )   1  ))   

Vertical stacking: [[1 2] [3 4] [5 6] [7 8]] Horizontal stacking: [[1 2 5 6] [3 4 7 8]] Column stacking: [[1 2 5] [3 4 6]] Concatenating to 2nd axis: [[1 2 5 6] [3 4 7 8]] 

np.hsplit: Jaa taulukko vaaka-akselia pitkin. np.vsplit: Jaa taulukko pystyakselia pitkin. np.array_split: Jaa taulukko määrättyä akselia pitkin.

   import   numpy   as   np   a   =   np  .  array  ([[  1     3     5     7     9     11  ]   [  2     4     6     8     10     12  ]])   # horizontal splitting   print  (  'Splitting along horizontal axis into 2 parts:  n  '     np  .  hsplit  (  a     2  ))   # vertical splitting   print  (  '  n  Splitting along vertical axis into 2 parts:  n  '     np  .  vsplit  (  a     2  ))   

Splitting along horizontal axis into 2 parts: [array([[1 3 5] [2 4 6]]) array([[ 7 9 11] [ 8 10 12]])] Splitting along vertical axis into 2 parts: [array([[ 1 3 5 7 9 11]]) array([[ 2 4 6 8 10 12]])] 

 A(2-D array): 4 x 3 B(1-D array): 3 Result : 4 x 3    
 A(4-D array): 7 x 1 x 6 x 1 B(3-D array): 3 x 1 x 5 Result : 7 x 3 x 6 x 5   But this would be a mismatch:  
 A: 4 x 3 B: 4   The simplest broadcasting example occurs when an array and a scalar value are combined in an operation. Consider the example given below: Python   
   import   numpy   as   np   a   =   np  .  array  ([  1.0     2.0     3.0  ])   # Example 1   b   =   2.0   print  (  a   *   b  )   # Example 2   c   =   [  2.0     2.0     2.0  ]   print  (  a   *   c  )   
  Output:  
[ 2. 4. 6.] [ 2. 4. 6.] 
 We can think of the scalar b being stretched during the arithmetic operation into an array with the same shape as a. The new elements in b as shown in above figure are simply copies of the original scalar. Although the stretching analogy is only conceptual. Numpy is smart enough to use the original scalar value without actually making copies so that broadcasting operations are as memory and computationally efficient as possible. Because Example 1 moves less memory (b is a scalar not an array) around during the multiplication it is about 10% faster than Example 2 using the standard numpy on Windows 2000 with one million element arrays! The figure below makes the concept more clear:    In above example the scalar b is stretched to become an array of with the same shape as a so the shapes are compatible for element-by-element multiplication. Now let us see an example where both arrays get stretched. Python   
   import   numpy   as   np   a   =   np  .  array  ([  0.0     10.0     20.0     30.0  ])   b   =   np  .  array  ([  0.0     1.0     2.0  ])   print  (  a  [:   np  .  newaxis  ]   +   b  )   
  Output:  
[[ 0. 1. 2.] [ 10. 11. 12.] [ 20. 21. 22.] [ 30. 31. 32.]]  
   Joissakin tapauksissa yleislähetys venyttää molempia matriiseja muodostaen lähtötaulukon, joka on suurempi kuin jompikumpi alkuperäisistä taulukoista.    Työskentely päivämäärän ja ajan kanssa:   Numpy has core array data types which natively support datetime functionality. The data type is called datetime64 so named because datetime is already taken by the datetime library included in Python. Consider the example below for some examples: Python   
   import   numpy   as   np   # creating a date   today   =   np  .  datetime64  (  '2017-02-12'  )   print  (  'Date is:'     today  )   print  (  'Year is:'     np  .  datetime64  (  today     'Y'  ))   # creating array of dates in a month   dates   =   np  .  arange  (  '2017-02'     '2017-03'     dtype  =  'datetime64[D]'  )   print  (  '  n  Dates of February 2017:  n  '     dates  )   print  (  'Today is February:'     today   in   dates  )   # arithmetic operation on dates   dur   =   np  .  datetime64  (  '2017-05-22'  )   -   np  .  datetime64  (  '2016-05-22'  )   print  (  '  n  No. of days:'     dur  )   print  (  'No. of weeks:'     np  .  timedelta64  (  dur     'W'  ))   # sorting dates   a   =   np  .  array  ([  '2017-02-12'     '2016-10-13'     '2019-05-22'  ]   dtype  =  'datetime64'  )   print  (  '  n  Dates in sorted order:'     np  .  sort  (  a  ))   
  Output:  
Date is: 2017-02-12 Year is: 2017 Dates of February 2017: ['2017-02-01' '2017-02-02' '2017-02-03' '2017-02-04' '2017-02-05' '2017-02-06' '2017-02-07' '2017-02-08' '2017-02-09' '2017-02-10' '2017-02-11' '2017-02-12' '2017-02-13' '2017-02-14' '2017-02-15' '2017-02-16' '2017-02-17' '2017-02-18' '2017-02-19' '2017-02-20' '2017-02-21' '2017-02-22' '2017-02-23' '2017-02-24' '2017-02-25' '2017-02-26' '2017-02-27' '2017-02-28'] Today is February: True No. of days: 365 days No. of weeks: 52 weeks Dates in sorted order: ['2016-10-13' '2017-02-12' '2019-05-22'] 
     Lineaarinen algebra NumPyssä: NumPyn Linear Algebra -moduuli tarjoaa erilaisia ​​​​menetelmiä lineaarisen algebran soveltamiseen mihin tahansa numpy-taulukkoon. Löydät: 
  
taulukon rank determinant trace jne.. 
  
omia arvoja tai matriiseja 
  
matriisi ja vektoritulot (pisteen sisäinen ulko jne. tulo) matriisin eksponentio 
  
ratkaise lineaarisia tai tensoriyhtälöitä ja paljon muuta! 
  
 Consider the example below which explains how we can use NumPy to do some matrix operations. Python   
   import   numpy   as   np   A   =   np  .  array  ([[  6     1     1  ]   [  4     -  2     5  ]   [  2     8     7  ]])   print  (  'Rank of A:'     np  .  linalg  .  matrix_rank  (  A  ))   print  (  '  n  Trace of A:'     np  .  trace  (  A  ))   print  (  '  n  Determinant of A:'     np  .  linalg  .  det  (  A  ))   print  (  '  n  Inverse of A:  n  '     np  .  linalg  .  inv  (  A  ))   print  (  '  n  Matrix A raised to power 3:  n  '     np  .  linalg  .  matrix_power  (  A     3  ))   
  Output:  
Rank of A: 3 Trace of A: 11 Determinant of A: -306.0 Inverse of A: [[ 0.17647059 -0.00326797 -0.02287582] [ 0.05882353 -0.13071895 0.08496732] [-0.11764706 0.1503268 0.05228758]] Matrix A raised to power 3: [[336 162 228] [406 162 469] [698 702 905]] 
 Let us assume that we want to solve this linear equation set:  
 x + 2*y = 8 3*x + 4*y = 18   This problem can be solved using   linalg.solve   method as shown in example below: Python   
   import   numpy   as   np   # coefficients   a   =   np  .  array  ([[  1     2  ]   [  3     4  ]])   # constants   b   =   np  .  array  ([  8     18  ])   print  (  'Solution of linear equations:'     np  .  linalg  .  solve  (  a     b  ))   
  Output:  
Solution of linear equations: [ 2. 3.] 
 Finally we see an example which shows how one can perform linear regression using least squares method. A linear regression line is of the form w1  x + w 2 = y ja se on viiva, joka minimoi kunkin datapisteen ja suoran välisen etäisyyden neliöiden summan. Kun siis annetaan n dataparia (xi yi), etsimämme parametrit ovat w1 ja w2, jotka minimoivat virheen:  Let us have a look at the example below: Python   
   import   numpy   as   np   import   matplotlib.pyplot   as   plt   # x co-ordinates   x   =   np  .  arange  (  0     9  )   A   =   np  .  array  ([  x     np  .  ones  (  9  )])   # linearly generated sequence   y   =   [  19     20     20.5     21.5     22     23     23     25.5     24  ]   # obtaining the parameters of regression line   w   =   np  .  linalg  .  lstsq  (  A  .  T     y  )[  0  ]   # plotting the line   line   =   w  [  0  ]  *  x   +   w  [  1  ]   # regression line   plt  .  plot  (  x     line     'r-'  )   plt  .  plot  (  x     y     'o'  )   plt  .  show  ()   
  Output: