Hi¡ I have a doubt and I hope someone can help me please, I have a dataframe in R and it makes a double cicle for and an if, the data frame has some values and then if the condition is True, it makes some operations, the problem is I can't understand neither the cicle and the operation the code makes under the condition.
I reply the code I have in a simpler one but the idea is the same. And if someone can explain me the whole operation please.
w<-c(2,5,4,3,5,6,8,2,4,6,8)
x<-c(2,5,6,7,1,1,4,9,8,8,2)
y<-c(2,5,6,3,2,4,5,6,7,3,5)
z<-c(2,5,4,5,6,3,2,5,6,4,6)
letras<-data.frame(w,x,y,z)
l=1
o=1
v=nrow(letras)
letras$op1<-c(1)
letras$op2<-c(0)
for (l in 1:v) {
for (o in 1:v) {
if(letras$x[o]==letras$y[l] & letras$z[l]==letras$z[o] & letras$w[l]){
letras$op1<-letras$op1+1
letras$op2<-letras$x*letras$y
}
}
}
The result is the following:
Thanks¡¡¡¡¡
This segment of code is storing values into vectors labeled w,x,y,z.
w<-c(2,5,4,3,5,6,8,2,4,6,8)
x<-c(2,5,6,7,1,1,4,9,8,8,2)
y<-c(2,5,6,3,2,4,5,6,7,3,5)
z<-c(2,5,4,5,6,3,2,5,6,4,6)
It then transforms the 4 vectors into a data frame
letras<-data.frame(w,x,y,z)
This bit of code isn't doing anything as far as I can tell.
l=1 #???
o=1 #???
This counts how many rows is in the letras data frame and stores to v, in this case 11 rows.
v=nrow(letras)
This creates new columns in letras dataframe with all ones in op1 and all zeros in op2
letras$op1<-c(1)
letras$op2<-c(0)
Here each for loop is acting as a counter, and will run the code beneath it iteratively from 1 to v (11), so 11 iterations. Each iteration the value of l will increase by 1. So first iteration l = 1, second l=2... etc.
for (l in 1:v) {
You then have a second counter, which is running within the first counter. So this will iterate over 1 to 11, exactly the same way as above. But the difference is, this counter will need to complete it's 1 to 11 cycle before the top level counter can move onto the next number. So o will effectively cycle from 1 to 11, for each 1 count of 1l. So with the two together, the inside for loop will count from 1 to 11, 11 times.
for (o in 1:v) {
You then have a logical statement which will run the code beneath if the column x and column y values are the same. Remember they will be calling different index values so it could be 1st x value vs the 2nd y value. There is an AND statement so it also needs the two z position values to be equal. and the last part letras$w[l] is always true in this particular example, so could possibly be removed.
if(letras$x[o]==letras$y[l] & letras$z[l]==letras$z[o] & letras$w[l]){
Lastly, is the bit that happens if the above statement is true.
op1 get's 1 added (remember this was starting from 1 anyway), and op2 multiplies x*y columns together. This multiplication is perhaps a little bit inefficient, because x and y do not change, so the answer will calculate the same result each time the the if statement evaluates TRUE.
letras$op1<-letras$op1+1
letras$op2<-letras$x*letras$y
}
}
}
Hope this helps.
Below is a MIPS program that converts from Fahrenheit to Celsius.
I've defined 3 variables, one of which is a double and the rest are all integer. In order to perform division or multiplication between the variables, however, I had to convert the int variables to double later in the code.
The whole reason I declared the variables as int not float or double is because I'm literally storing one-digit decimal constants, and I wouldn't need to use two registers for that, asdouble requires. However, now I realise I'm facing two new problems:
1) I now need to do two extra steps for each variable (moving to cp1 and converting to double)
2) After the line mtc1.d $t1, $f6, the values become double and they get stored in two registers like normal doubles instead of one register like integers, hence missing my intention completely.
My question is, which way is better for memory and program efficiency? Declaring the variables as all of the same data type and shortening the number of instructions but taking up more memory space than I need?
Or giving the variables more suitable data types but costing myself 3 steps instead of just one?
Isn't there an efficient way to do arithmetic between two different data types in MIPS by using as few registers as possible?
.data
input: .double
cons1: .double 32
cons2: .word 5
cons3: .word 9
.text
#take double input from user
li $v0, 7
syscall #value is stored in f0 (and f1 probably)
#subtract first constant
ldc1 $f2, cons1 #load the value 32 into f2 and f3
sub.d $f12, $f0, $f2
#multiply by 5
lw $t1, cons2 #load 5 into t1
mtc1.d $t1, $f4 #move the value in t1 to f4
cvt.d.w $f4, $f4 #convert value to single precision to be able to do
#multiplication
mul.d $f12, $f12, $f4
#divide by 9
lw $t2, cons3 #load 9 into t2
mtc1.d $t2, $f6 #move the value in t2 to f5
cvt.d.w $f6, $f6 #convert value to single precision to be able to do
#division
div.d $f12, $f12, $f6
#show result
li $v0, 3
syscall
I've been working on a hex calculator for a while, but seem to be stuck on the subtraction portion, particularly when B>A. I'm trying to simply subtract two positive integers and display the result. It works fine for A>B and A=B. So far I'm able use two 7-segment displays to show the integers to be subtracted and I get the proper difference as long as A>=B
When B>A I see a pattern that I'm not able to debug because of my limited knowledge in Verilog case/if-else statements. Forgive me if I'm not explaining the best way but what I'm observing is that once the first number, A, "reaches" 0 (after being subtracted from) it loops back to F. The remainder of B is then subtracted from F rather than 0.
For example: If A=1, B=3
A - B =
1 - 1 = 0
0 - 1 = F
F - 1 = E
Another example could be 4-8=C
Below are the important snippets of code I've put together thus far.
First, my subtraction statement
always#*
begin
begin
Cout1 = 7'b1000000; //0
end
case(PrintDifference[3:0])
4'b0000 : Cout0 = 7'b1000000; //0
4'b0001 : Cout0 = 7'b1111001; //1
...
4'b1110 : Cout0 = 7'b0000110; //E
4'b1111 : Cout0 = 7'b0001110; //F
endcase
end
My subtraction is pretty straightforward
output [4:0]Difference;
output [4:0] PrintDifference;
assign PrintDifference = A-B;
I was thinking I could just do something like
if A>=B, Difference = B-A
else, Difference = A-B
Thank you everyone in advance!
This is expected behaviour of twos complement addition / subtraction which I would recommend reading up on since it is so essential.
The result obtained can be changed back into an unsigned form by inverting all the bits and adding one. Checking the most significant bit will tell you if the number is negative or not.
I know this algorithm is for finding the majority element of any array if it has any. Can any one please explain the recursion calls?
if length(A) = 0 then
return null
end if
if length(A) = 1 then
return 1
end if
// "Command 7"
Call FIND-MAJORITY recursively on the first half of A, and let i be the result.
// "Command 8"
Call FIND-MAJORITY recursively on the second half of A, and let j be the result.
if i > 0 then
compare i to all objects in A(including itself);
let k be the number of times that equality holds;
if k > length(A)/2 then
return i.
end if
end if
if j > 0 then
compare j to all objects in A(including itself);
let k be the number of times that equality holds;
if k > length(A)/2 then
return j
end if
end if
return null
Is command 7 is executed until it get an single value ... and then command 8? I cannot understand these recursions. Please explain with example, thanks.
It depends what are inputs of this function.
If the array A is an input then we only search in the diminished array else if the array A is defined as global then you always search the whole array.
For example take the array A is 1,2,1,3,1,8,7,1
If the array is given as input to the function :
According to recursion we get A is 1,2,1,3 -> A is 1,2 -> A is 1
This returns i := 1.
Then A is 2, this returns j:=1.
Then we compare i to all elements of A i.e 1,2.
Then we compare j to all elements of A i.e. 1,2.
We return null from this recursive call.
After this we proceed to upper recursion i.e. 1,2,1,3 and up to the first call.
If the array is global:
According to recursion we get A is 1,2,1,3 -> A is 1,2 -> A is 1
This returns i := 1.
Then A is 2, this returns j:=1.
Then we compare i to all elements of A i.e. 1,2,1,3,1,8,7,1
we return according to the conditions.
**Remember even in this case we return all recursive calls and check the whole array for every recursive call which is not what you probably want.
So, I am working on a program in Scilab which solves a binary puzzle. I have come across a problem however. Can anyone explain to me the logic behind solving a binary sequence with gaps (like [1 0 -1 0 -1 1 -1] where -1 means an empty cell. I want all possible solutions of a given sequence. So far I have:
function P = mogelijkeCombos(V)
for i=1:size(V,1)
if(V(i) == -1)
aantalleeg = aantalleeg +1
end
end
for i=1:2^aantalleeg
//creating combos here
end
endfunction
sorry that some words are in dutch
aantalleeg means amountempty by which I mean the amount of empty cells
I hope I gave you guys enough info. I don't need any code written, I'd just like ideas of how I can make every possible rendition as I am completely stuck atm.
BTW this is a school assignment, but the assignment is way bigger than this and it's just a tiny part I need some ideas on
ty in advance
Short answer
You could create the combos by extending your code and create all possible binary words of the length "amountempty" and replacing them bit-for-bit in the empty cells of V.
Step-by-step description
Find all the empty cell positions
Count the number of positions you've found (which equals the number of empty cells)
Create all possible binary numbers with the length of your count
For each binary number you generate, place the bits in the empty cells
print out / store the possible sequence with the filled in bits
Example
Find all the empty cell positions
You could for example check from left-to-right starting at 1 and if a cell is empty add the position to your position list.
V = [1 0 -1 0 -1 1 -1]
^ ^ ^
| | |
1 2 3 4 5 6 7
// result
positions = [3 5 7]
Count the number of positions you've found
//result
amountempty = 3;
Create all possible binary numbers with the length amountempty
You could create all possible numbers or words with the dec2bin function in SciLab. The number of possible words is easy to determine because you know how much separate values can be represented by a word of amountempty bits long.
// Create the binary word of amountEmpty bits long
binaryWord = dec2bin( i, amountEmpty );
The binaryWord generated will be a string, you will have to split it into separate bits and convert it to numbers.
For each binaryWord you generate
Now create a possible solution by starting with the original V and fill in every empty cell at the position from your position list with a bit from binaryWordPerBit
possibleSequence = V;
for j=1:amountEmpty
possibleSequence( positions(j) ) = binaryWordPerBit(j);
end
I wish you "veel succes met je opdracht"