Difference of Execution in Function and Without Function (Return Variable)

Question

So I have to following code with the goal to add some offset to measurement data.

This works as planned:

IndexMax = [3, 10, 20, 30]
IndexMax2 = [3, 10, 20, 30] #Just for Computing purposes


for x in range(0, len(IndexMax)):
  if x == 0:
      IndexMax2[0] = 0; #First Data has no Offset per Definition
  elif len(IndexMax2) >= 1: #The 
      IndexMax2[x] = IndexMax[x-1] + IndexMax2[x-1]
      tmp = IndexMax2


print('Offsets are ' + str(tmp))

The Correct Output is given:

Offsets are [0, 3, 13, 33]

But what does not work is the same as a funtion with Indexmax as an argument:

def Offsetcalcalculater(IndexMax):
     IndexMax = IndexMax
     IndexMax2 = IndexMax


     for x in range(0, len(IndexMax)):
         if x == 0:
              IndexMax2[0] = 0; #First Data has no Offset per Definition
         elif len(IndexMax2) >= 1: #The 
              IndexMax2[x] = IndexMax[x-1] + IndexMax2[x-1]
              tmp = IndexMax2
     return IndexMax2

Calling it with

IndexMaxtest = [3, 10, 20, 30]
IndexMax2 = Offsetcalcalculater(IndexMaxtest)

leads to:

IndexMax2 [0, 0, 0, 0]

Why does the function not behave in the same way as the code above?

Epsi95 · Accepted Answer · 2021-07-07 17:01:40Z

1

You are doing a shallow copy of IndexMax Both IndexMax and IndexMax2 referring same list

IndexMax    Indexmax2
#    |            |
#    |            |
#     \          /  
#      \        /
#       IndexMax

To solve this use .copy to make a copy of the list.

def Offsetcalcalculater(IndexMax):
     IndexMax = IndexMax
     IndexMax2 = IndexMax.copy() #---->


     for x in range(0, len(IndexMax)):
         if x == 0:
              IndexMax2[0] = 0; #First Data has no Offset per Definition
         elif len(IndexMax2) >= 1: #The 
              IndexMax2[x] = IndexMax[x-1] + IndexMax2[x-1]
              tmp = IndexMax2
     return IndexMax2

answered Jul 7, 2021 at 17:01

Epsi95

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Comments

Yevhen Kuzmovych · Accepted Answer · 2021-07-07 17:01:48Z

0

IndexMax = [3, 10, 20, 30]
IndexMax2 = [3, 10, 20, 30]

are two references to two different lists. If you change one, the second one won't change.

 IndexMax = IndexMax
 IndexMax2 = IndexMax

on the other hand, are two references to the same list. Changing something in IndexMax will lead to the same change in IndexMax2

answered Jul 7, 2021 at 17:01

Yevhen Kuzmovych

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