return a list for the given number-CodePudding

I know how assign items to empty array using append :

matri=[]
def matr_meth(n):
 
  for i in range(1,n 1):
    matri.append(i)
  return matri  

print(matr_meth(6))

but here I saw another way that I don't get how can we assign the index to the name of array matri =[i] , I would get it if matri[i] =i 1 for exaple cuz alawys array is compared to index like c and c maybe , second in the new method it won't work if I define the matri=[] before the method same as I did in the append way:

def matr_meth(n):
  matri=[]
  for i in range(1,n 1):
    matri  =[i]
  return matri  

print(matr_meth(6))

CodePudding user response：

You've stumbled upon two related but distinct questions here. Let's start with the first one. [i], on its own, is not an array subscript operator. It's a list with one element in it. So

matri  = [i]

is really just

matri = matri   [i]

or, more verbosely,

my_temporary_list = [i]
matri = matri   my_temporary_list

and is defined on lists to be concatenation. So it's just a fancier way of putting the two lists together. It's not really useful in this example (I would definitely use .append if I only had one element to append), but if you have a whole list (whose length might be greater than one), it's quicker to append it in one go. For instance, your for loop

for i in range(1,n 1):
  matri  =[i]

can actually be written as

matri  = range(1, n   1)

= on list can take any iterable, such as range, as an argument.

Now for your second question.

def matr_meth(n):
  matri=[]
  for i in range(1,n 1):
    matri  = [i]
  return matri

Here, matri is a local variable in matr_meth.

matri=[]
def matr_meth(n):
  for i in range(1,n 1):
    matri.append(i)
  return matri

Here, matri is a global variable, plain and simple. Where it gets nebulous is

matri=[]
def matr_meth(n):
  for i in range(1,n 1):
    matri  = [i]
  return matri

Now we're assigning to matri both inside and outside of the function (remember, = is just a shortcut for = and together). So Python could reasonably assume that matri is a global variable, or it could reasonably assume that you meant to make a new local variable inside the function. To avoid action-at-a-distance errors, Python makes the conservative choice to assume you meant matri to be a new local variable. Since you intended the opposite in this case, just use the global keyword.

matri=[]
def matr_meth(n):
  global matri
  for i in range(1,n 1):
    matri  = [i]
  return matri

CodePudding user response：

Lists in Python support the addition operator:

>>> [1, 2, 3]   [4, 5]
[1, 2, 3, 4, 5]

With that loop you're effectively extending the matri list with single-element lists, i.e.:

| i | matri           | [i] | matri   [i]                  |
| - | --------------- | --- | ---------------------------- |
| 1 | []              | [1] | []   [1]                     |
| 2 | [1]             | [2] | [1]   [2]                    |
| 3 | [1, 2]          | [3] | [1, 2]   [3]                 |
...
| 6 | [1, 2, 3, 4, 5] | [6] | [1, 2, 3, 4, 5]   [6]        |

CodePudding user response：

def matr_meth(n):
  matri=[]
  for i in range(1,n 1):
    matri  = [i]
  return matri

You initialized matri=[] as an empty list, which is a list. [i] is also a list. If you apply the operator between two lists, it is essentially appending the second list to the first list.

In the for loop you are appending every [i] to matri. matri keeps changing in the for loop. You can view it by using the print function.

def matr_meth(n):
  matri=[]
  for i in range(1,n 1):
    matri  = [i] 
    print('At the ' str(i) '-th iteration, matri is: ' )
    print(matri)      
  return matri  


matr_meth(6)

Output:

At the 1-th iteration, matri is: 
[1]
At the 2-th iteration, matri is: 
[1, 2]
At the 3-th iteration, matri is: 
[1, 2, 3]
At the 4-th iteration, matri is: 
[1, 2, 3, 4]
At the 5-th iteration, matri is: 
[1, 2, 3, 4, 5]
At the 6-th iteration, matri is: 
[1, 2, 3, 4, 5, 6]