Python CookBook2 Fourth Python tips-return If an element exists in the list && create a list of lists without sharing references

If an element exists in the list, the returned

Task:

You have a list L, there is an index number I, if I is a valid index, return l[i], if not, return the default value V

Solution:

1. The
2. list supports bidirectional indexing, so I can be negative
   

    >>> def  list_get (L,i,v=none):  if -len (l) <= i < Len (l):  return   L[i]  else  :  return   v  >>> list_get ([1,2,3,4,5,6],3 4 

    

4. Exception mechanism
   

   def list2_get (l,i,v=None):    try:        return  l[i]     except indexerror:         return ' ERror '    >>> List2_get ([1,2,3,4,5,6],7)'ERror'

Create a list of lists without sharing references

Task:

Create a multi-dimensional list to avoid implicit reference sharing at the same time.

Solution:

Use list derivation to create a 5 x 10 matrix of all 0:

 for inch  for  in range (ten)>>> test_list[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]

Python CookBook2 Fourth Python tips-return If an element exists in the list && create a list of lists without sharing references