I am trying to learn the merge sort alogrithm but I am running in to a block in my understanding.
Follwoing is the starting piece of the merge sort alogrithm:
def mergeSort(nlist):
print("Splitting ",nlist)
if len(nlist)>1:
mid = len(nlist)//2
lefthalf = nlist[:mid]
righthalf = nlist[mid:]
mergeSort(lefthalf)
mergeSort(righthalf)
The function mergesort is being called over and over agin and this will change the value of the left half and right half so that when the second mergesort function is called it will have ony a single element. where does the rest of the list get saved and how does it get retrieved again ? . I am not able to understand the flow of the program.
How do I understand this I am not able to earp my head around how this works...
Related
I am fairly new to using PeopleSoft BI Publisher plugin for MS Word and integrating it with PS Query Manager. My question is whether in the RTF file you can put logic to suppress or filter out data?
I have a for-each grouping that prints a line (row). I would like to add logic to NOT print the line if the Witholding amount field (M.WTHD_AMT) is equal to 0 (zero). My question is what would the syntax look like, and where should I place it (on the For Each grouping below, the Field level, or somewhere else?) I know I can alter the PS Query (data source) to do the filtering but I would like to leave that as-is and handle this in the template.
I see that there is another conditional IF statement ("rmt_") so I'm not sure if I can add this additional logic to that element or if I need a separate one. I appreciate any feedback!
EDIT:
I've added a new "Conditional Region" as suggested, and it works with just the WTHD_AMT criteria !0 to zero, however I tried added additional criteria where L.PYMNT_TYPE = 'R' and when I run the process it doesn't display data on the PDF output. Is there something wrong with the syntax? Do I need to have a separate Conditional Region for this 2nd criteria? I've seen another BI report where they have 2 or 3 criteria as part of one element.
<?if:number(M.WTHD_AMT)!=0.00?> and <?if:L.PYMNT_TYPE='R'?>
Option 1
You can nest <?if?> statements. Just add another <?end if?> at the end. Make sure there are no spaces between the all of the IF or END IF objects at the beginning or end of the content/row, else the row may still be displayed.
Option 2
You can add conditions in the repeating section. Below will repeat the region for every record where M.WTHD_AMT is not 0.00
<?for-each:record_path/record[M.WTHD_AMT!='0.00']?>
'Conditional Region' is the button you are looking for.
When using this button, make sure to double check where the if/endif or C/EC elements are added. It tends to ignore the selected element and join the elements to the start and end of the line. You will then need to cut and paste it into the right spot. For you this will probably be right after the F element and before the E element.
After studying the merge sort for a couple of days, I understand it conceptually, but there is one thing that I don't get.
What I get:
1.) It takes a list, for example an array of numbers and splits it in half and sorts the two halfs, and in the end merges them together.
2.) Because it's an recursive algorithm it uses recursion to do that.
So the split of the mentioned array looks like this:
It, splits the array until there is only one item in each list and by that its considered sorted. And at that point the merge steps in.
Which should look like this:
What I don't get is, how does the recursion "know" after it splits all the lists to only one item in a list, to get back up the recursion tree? How does something that has a left and right side become the left side after it merges?
The thing that bothers me is this. I've taken a snapshot of the code from interactivepython page
How does the code get to the point, after we have lefthalf = 2, and righthalf = 1, to to code that's shown in the picture where the lefthalf = [1,2] and righthalf = [4,3] without going back to the recursion that would divide what we have have merged?
Tnx,
Tom
Once the list only contains one element, each pair of leaves are sorted and joined. Then you can traverse through the list and find out where the next pair should be inserted. The recursion "knows" nothing about going back up the recursion tree, rather it is the act of sorting and joining that has this effect.
The "recursion" does of course know nothing of that sort. It is the code that uses the recursion, which looks like this (a bit simplified):
sort list = merge (sort left_half) (sort right_half)
where
(left_half, right_half) = split list
Here you see that the "recursion" (i.e. the recursive invocations of sort) don't need to "know" anything. Their only job is to deliver a sorted list, array or whatever.
To put it differently: If we have merge satisfying the following invariant:
1. `merge`, given two sorted lists, will return a sorted list.
then we can write mergesort easily like outlined above. What is left to do in sort is to handle the easy cases: empty list, singleton and list with two elements.
If you are talking about odd numbered sub lists, then it is dependant on the implementation.
It either puts the bigger sub list on the left every single time, or it puts it on the right every single time.
Hi I need to write a Prolog predicate that will sum the lines of a list of list so given
|a,b,c| |j,k,l| |s,t,u|
|d,e,f| |m,n,o| |v,w,x|
|g,h,i|.|p,q,r|,|y,z,?|
/*I want to be able to get back */
|(a+j+s),(b+k+t),(c+l+u)|
|(d+m+v),(e+n+w),(f+o+x)|
|(g+p+y),(h+q+z),(i+r+?)|
Im having a hard time because I want to add the head to the head of the tail, and so on but can't figure out away to tell prolog to do that. I can add any one line at a time or even just two matrices but given a list like this I cannot see to get it. Any help would be appreciated. Thanks.
I wrote a function in R - called "filtre": it takes a dataframe, and for each line it says whether it should go in say bin 1 or 2. At the end, we have two data frames that sum up to the original input, and corresponding respectively to all lines thrown in either bin 1 or 2. These two sets of bin 1 and 2 are referred to as filtre1 and filtre2. For convenience the values of filtre1 and filtre2 are calculated but not returned, because it is an intermediary thing in a bigger process (plus they are quite big data frame). I have the following issue:
(i) When I later on want to use filtre1 (or filtre2), they simply don't show up... like if their value was stuck within the function, and would not be recognised elsewhere - which would oblige me to copy the whole function every time I feel like using it - quite painful and heavy.
I suspect this is a rather simple thing, but I did search on the web and did not find the answer really (I was not sure of best key words). Sorry for any inconvenience.
Thxs / g.
It's pretty hard to know the optimum way of achieve what you want as you do not provide proper example, but I'll give it a try. If your variables filtre1 and filtre2 are defined inside of your function and you do not return them, of course they do not show up on your environment. But you could just return the classification and make filtre1 and filtre2 afterwards:
#example data
df<-data.frame(id=1:20,x=sample(1:20,20,replace=TRUE))
filtre<-function(df){
#example function, this could of course be done by bins<-df$x<10
bins<-numeric(nrow(df))
for(i in 1:nrow(df))
if(df$x<10)
bins[i]<-1
return(bins)
}
bins<-filtre(df)
filtre1<-df[bins==1,]
filtre2<-df[bins==0,]
I'm trying to create a histogram of an image. I was thinking to first bubblesort the array of the pixels so every number is sorted from low to high.
Then its easier to count how many times a specific value of a pixels appears. And then later I can put it in a graph.
But it always gives an error then I don't understand.
I also want to make everything with the formula node instead of just blocks.
Visual:
http://i.stack.imgur.com/ZlmW2.png
Error:
http://i.stack.imgur.com/91TbS.png
In your code numbers is a scalar not an array.
Besides that the formula node does not maintain state, you'll need a feedback node to get history. Is there any reason why do you want to use the formula node instead of native LabVIEW code?
You need to remove the two nested LabVIEW for loops, you are iterating through your array inside the formula node so you don't need to do it with the loops.