#Matlab for loop backwards code#

In this case the loop will continue to loop while index satisfies the logical statement here. I need to solve for C (n, a) which has to be solved backwards using A and B.

This type of loop will ALWAYS, and I repeat ALWAYS execute at least once. In the above example the loop will continue to loop until the index value index becomes greater than or equal to someValue at the end of the current iteration of the loop.

Many times you'll see this done with a while loop or a do while combo.Ī few examples are as follows: index = 0 Generally this isn't the best idea when you're using a for loop as the value of your indexing variable will be changed by the loop as well as by you within the loop. Input: for b 1:10 fprintf (‘value of b: d ’, b) end.

#Matlab for loop backwards code#

Secondly here you can modify the value within the loop itself. If expression is a range, a row vector, or a scalar, the value of var will be a scalar each time the loop body is executed. initval:endval: This value function can be used to increase the index variable that is incremented by 1 from initval to endval, and this will repeat the execution of the code statements until the index has greater value than endval. Namely trying to index out of bounds of the matrix or potentially going negative. You just have to be aware of potential indexing issues that can arrise. Obviously you can replace the value between the two : to be a step size that you want. It would look like this: for index = someValue:-1:1 You can also do every 4th index while going backwards for index = someValue:-4:1. In this case you start at the 'end' and move to the beginning. Yet another modfication is traversing backwards. All you have to do is: for index = 2:3:someValue The 3 here tells the loop that it should add 3 to the index at the end of each loop iteration until you get to (or surpass) someValue. You can modify this as well though! Let's say you want every 3rd index starting with the 2nd position in the matrix. Many times it's simply the last index value of a matrix you're wanting to traverse. SomeValue can really be anything that is greater than 1 in this case. The general syntax is for index = 1:someValue I will also open this up for community wiki so others can just easily edit this as well. There are two main ways that I can think of off the top of my head. Negative step values traverse backwards.Generally you and iterate multiple ways. Here is a way to reverse a string without utilizing the built in features such as reversed. The reverse_string code is almost identical to the PEP with a check removed for simplicity's sake. Put the value in M (i, j) location and it will insert the value in the matrix. Here row 5 and then column 3 and for hence two for loop. Or calling a constructor like: cba = list(reverse_string('abc')) To create and add a value into the matrix you can do this and can make a complete matrix like yours. Without using the builtin, it might look something like, def reverse_string(x: str) -> str:Ĭonsume the iterator either by looping, eg for element in (reverse_string('abc')): So to reverse a string, consume the iterator until it is exhausted. makes a reverse iterator over sequence objects that support getitem() and len(). The function checks whether the argument is iterable and then yields the last element of a list until there are no more elements to yield. For those readers attempting to understand how reversed is implemented, take a look at the PEP, number 322, to get an understanding of the how and why. But this is error prone and the builtin function reversed is a much better approach. For Loop Backwards Help Verfolgen 511 Ansichten (letzte 30 Tage) Ältere Kommentare anzeigen WhatIsMatlab- am 7 Feb. It does show how one could utilize range and negative step values to build a value by looping through a string and adding elements in off the end of the string to the front of the new value. It is not a very pythonic or even efficient way to loop over a string backwards. Best Answer Here you go: x zeros ( 1, 11) for i 11:-1:2 x (i-1) x (i) + ( (5 - (4/50)x (i))) end display (x)x 35.351 32.99 30.424 27.635 24.603 21.307 17.725 13.832 9. 1 Answer Sorted by: 2 Syntax As the for-loop documentation states, the for loop's syntax is: for variableexpression do instruction, ,instruction,end If expression is a matrix or a row vector, variable takes as values the values of each column of the matrix. EDIT: It has been quite some time since I wrote this answer.