What is the difference between i i 1 and i 1 in a for loop

Question

I found out a curious thing today and was wondering if somebody could shed some light into what the difference is here   import numpy as np  A   np arange 12  reshape 4 3  for a in A      a   a   1  B   np arange 12  reshape 4 3  for b in B      b    1   After running each for loop  A has not changed  but B has had one added to each element  I actually use the B version to write to a initialized NumPy array within a for loop

User · Answer

In the first example  you are reassigning the variable a  while in the second one you are modifying the data in-place  using the    operator   See the section about 7 2 1  Augmented assignment statements       An augmented assignment expression like x    1 can be rewritten as x   x   1 to achieve a similar  but not exactly equal effect  In the augmented version  x is only evaluated once  Also  when possible  the actual operation is performed in-place  meaning that rather than creating a new object and assigning that to the target  the old object is modified instead       operator calls   iadd    This function makes the change in-place  and only after its execution  the result is set back to the object you are  applying  the    on     add   on the other hand takes the parameters and returns their sum  without modifying them

User · Answer

A key issue here is that this loop iterates over the rows  1st dimension  of B   In  258   B Out 258    array    0   1   2            3   4   5            6   7   8            9  10  11    In  259   for b in B                print b    gt   end                   b    1               print b                  0 1 2    gt  1 2 3   3 4 5    gt  4 5 6   6 7 8    gt  7 8 9    9 10 11    gt  10 11 12    Thus the    is acting on a mutable object  an array   This is implied in the other answers  but easily missed if your focus is on the a   a 1 reassignment   I could also make an in-place change to b with     indexing  or even something fancier  b 1   0   In  260   for b in B                print b    gt   end                   b      b   2   1 2 3    gt  2 4 6   4 5 6    gt   8 10 12   7 8 9    gt  14 16 18   10 11 12    gt  20 22 24    Of course with a 2d array like B we usually don t need to iterate on the rows   Many operations that work on a single of B also work on the whole thing   B    1  B 1     0  etc

User · Answer

First off  The variables a and b in the loops refer to numpy ndarray objects   In the first loop  a   a   1 is evaluated as follows  the   add   self  other  function of numpy ndarray is called  This creates a new object and hence  A is not modified  Afterwards  the variable a is set to refer to the result    In the second loop  no new object is created  The statement b    1 calls  the   iadd   self  other  function of numpy ndarray which modifies the ndarray object in place to which b is referring to  Hence  B is modified

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The difference is that one modifies the data-structure itself  in-place operation  b    1 while the other just reassigns the variable a   a   1     Just for completeness   x    y is not always doing an in-place operation  there are  at least  three exceptions    If x doesn t implement an   iadd   method then the x    y statement is just a shorthand for x   x   y  This would be the case if x was something like an int  If   iadd   returns NotImplemented  Python falls back to x   x   y  The   iadd   method could theoretically be implemented to not work in place  It d be really weird to do that  though    As it happens your bs are numpy ndarrays which implements   iadd   and return itself so your second loop modifies the original array in-place   You can read more on this in the Python documentation of  Emulating Numeric Types       These    i     methods are called to implement the augmented arithmetic assignments      -                              lt  lt     gt  gt     amp             These methods should attempt to do the operation in-place  modifying self  and return the result  which could be  but does not have to be  self   If a specific method is not defined  the augmented assignment falls back to the normal methods  For instance  if x is an instance of a class with an   iadd     method  x    y is equivalent to x   x   iadd   y    Otherwise  x   add   y  and y   radd   x  are considered  as with the evaluation of x   y  In certain situations  augmented assignment can result in unexpected errors  see Why does a tuple i       item   raise an exception when the addition works    but this behavior is in fact part of the data model

User · Answer

As already pointed out  b    1 updates b in-place  while a   a   1 computes a   1 and then assigns the name a to the result  now a does not refer to a row of A anymore    To understand the    operator properly though  we need also to understand the concept of mutable versus immutable objects  Consider what happens when we leave out the  reshape   C   np arange 12  for c in C      c    1 print C       0  1  2  3  4  5  6  7  8  9 10 11    We see that C is not updated  meaning that c    1 and c   c   1 are equivalent  This is because now C is a 1D array  C ndim    1   and so when iterating over C  each integer element is pulled out and assigned to c   Now in Python  integers are immutable  meaning that in-place updates are not allowed  effectively transforming c    1 into c   c   1  where c now refers to a new integer  not coupled to C in any way  When you loop over the reshaped arrays  whole rows  np ndarray s  are assigned to b  and a  at a time  which are mutable objects  meaning that you are allowed to stick in new integers at will  which happens when you do a    1   It should be mentioned that though   and    are meant to be related as described above  and very much usually are   any type can implement them any way it wants by defining the   add   and   iadd   methods  respectively

User · Answer

The short form a    1  has the option to modify a in-place   instead of creating a new object representing the sum and rebinding it back to the same name a   a   1  So The short form a    1  is much efficient as it doesn t necessarily need to make a copy of a unlike a   a   1   Also even if they are outputting the same result  notice they are different because they are separate operators    and

[python] What is the difference between i = i + 1 and i += 1 in a 'for' loop?

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