I have these ranges:
X = [10..17,21..34,40..117,384..399,407..455,989..1013]
What I need is:
if the first number of the second range minus the last number of the first range is less than 10, then group them together, else include the range.
What I need is the following:
result = [[10..17,21..34,40..117],[384..399,407..455],[989..1013]]
what I've done is this but this does not group more than two together.
Y = []
X.each_cons(2) do |r1,r2|
if (r2.first - r1.last) < 10
Y << ((r1.last - r1.first) + (r2.last - r2.first)
else
Y << (r1.last - r1.first)
end
end
# Y = [[10..17,21..34,40..117],[384..399,407..455],[989..1013]]
I like to also then add up all ranges within each element:
Y = [97, 63, 24]
Any help will be appreciated. Thanks in advance.
prev = X.first.last
# => 17
X.slice_before{|r| (r.first - prev >= 10).tap{prev = r.last}}.to_a
# => [[10..17, 21..34, 40..117], [384..399, 407..455], [989..1013]]
Related
I am working on an algorithms problem. You have an array numbers, size of array t , number number_of_elements and number multiplication_value. You have to find any set of number_of_elements indexes of the elements of the array , which product will be equal to multiplication_value. It is guaranteed, that such set of indexes exists
That problem looks like 2 sum, but I can't extrapolate it to my case.
I have tried naive algorithm for O(n), but it fails, when you have bad first number in an array. I think there is a way to use recursion in here. I guess it is well-known problem, but I couldn't find the solution
Example in:
t = 7
number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]
Example out:
1 3
My code ideas:
def return_index_values(numbers,multiplication_value,number_of_elements):
cur_number = int(multiplication_value)
list_of_indexes = []
values = []
for i in range(len(numbers)):
if ((cur_number == 1) and (len(values) == number_of_elements)):
print(values)
#finishing if everything worked
break
else:
if (cur_number % int(numbers[i]) == 0):
if(len(values) < number_of_elements):
#pushing values if possible
values.append(int(numbers[i]))
list_of_indexes.append(i)
cur_number = int(cur_number / int(numbers[i]))
print(cur_number)
else:
pass
if(len(values) == number_of_elements):
if mult_check(values,int(multiplication_value)):
#mult_check checks if the array's element multiplication gives a value
break
else:
#started dealing with bad cases, but it doesn't work properly
values.sort()
val_popped = values.pop()
cur_number = cur_number * val_popped
Bad case for my code
numbers = [9,3,1,27,3,27,3]
Here is one implementation. Not necessarily the best solution but it gives you some sense of how it can be done.
It first sorts the numbers by the element keeping the indices information. Then it performs recursion calls.
number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]
def preprocess(numbers, multiplication_value, number_of_elements):
l = []
for i, num in enumerate(numbers):
l.append((num, i))
return sorted(l, key = lambda tup: tup[0])
def subroutine(numbers, multiplication_value, number_of_elements, idx_start, result):
if idx_start >= len(numbers):
return False
if number_of_elements == 0:
return True if multiplication_value == 1 else False
for i in range(idx_start, len(numbers)):
num = numbers[i][0]
if num <= multiplication_value:
if multiplication_value % num == 0:
idx = numbers[i][1]
result.append(idx)
found = subroutine(numbers, multiplication_value / num, number_of_elements - 1, i + 1, result)
if not found:
del result[-1]
else:
return True
else:
return False
return False
result = []
processed_numbers = preprocess(numbers, multiplication_value, number_of_elements)
subroutine(processed_numbers, multiplication_value, number_of_elements, 0, result)
print(result)
You can use itertools.combinations() (https://www.geeksforgeeks.org/itertools-combinations-module-python-print-possible-combinations/) to select number_of_elements entries from your list in all possible ways, then check each whether they multiply to the required number.
puts "Enter range(starts at 1), ends at the number that you enter: "
range = gets.chomp.to_i
number = 1
while number <= range
temporary_number = number
sum_angstrom = 0
number += number
while(temporary_number != 0)
digit = temporary_number % 10
temporary_number /= 10
sum_angstrom = sum_angstrom + (digit ** 3)
end
if (sum_angstrom == number)
puts number
end
end
This time, I tried to make a program to show the armstrong numbers in a range that's taken from the user's input. The program just stops after I enter the number and press enter and i can't figure out why.
Keep in mind that i can't use for(each), that's why i'm using while so often.
First of all, change number += number to number += 1; otherwise you will only test the powers of 2.
Second, move the number += 1 line at the bottom of the while block it is in. Otherwise you will always test if sum_armstrong(n) == n+1.
This works:
puts "Enter range(starts at 1), ends at the number that you enter: "
range = gets.chomp.to_i
number = 1
while number <= range
temporary_number = number
sum_angstrom = 0
while(temporary_number != 0)
digit = temporary_number % 10
temporary_number /= 10
sum_angstrom = sum_angstrom + (digit ** 3)
end
if (sum_angstrom == number)
puts number
end
number += 1
end
Armstrong Number in Ruby one liner
n = 153
s = 0
n.to_s.split("").map{|e| s+=(e.to_i*e.to_i*e.to_i)}
puts (n==s ? "Armstrong number" : "Not Armstrong number")
You can iterate in a range to print the value based on your requirement.
Main logic lies in below line.
n.to_s.split("").map{|e| s+=(e.to_i*e.to_i*e.to_i)}
Improving my answer a little bit
n.digits.map{|e| s+=(e**3)}
I'm really struggling to design an algorithm to find d, which is the lowest value that can be added or subtracted (at most) to make a given sequence strictly increasing.
For example.. say seq[] = [2,4,8,3,1,12]
given that sequence, the algorithm should return "5" as d because you can add or subtract at most 5 to each element such that the function is strictly increasing.
I've tried several approaches and can't seem to get a solid technique down.
I've tried looping through the seq. and checking if seq[i] < seq[i+1]. If not, it checks if d>0.. if it is, try to add/subtract it from seq[i+1]. Otherwise it calculates d by taking the difference of seq[i-1] - seq[i].
I can't get it to be stable though and Its like I keep adding if statements that are more "special cases" for unique input sequences. People have suggested using a binary search approach, but I can't make sense of applying it to this problem.
Any tips and suggestions are greatly appreciated. Thanks!
Here's my code in progress - using Python - v4
def ComputeMaxDelta3(seq):
# Create a copy to speed up comparison on modified values
aItems = seq[1:] #copies sequence elements from 1 (ignores seq[0])
# Will store the fix values for every item
# this should allocate 'length' times the 0 value
fixes = [0] * len(aItems)
print("fixes>>",fixes)
# Loop until no more fixes get applied
bNeedFix = True
while(bNeedFix):
# Hope will have no fix this turn
bNeedFix = False
# loop all subsequent item pairs (i should run from 0 to length - 2)
for i in range(0,len(aItems)-1):
# Left item
item1 = aItems[i]
# right item
item2 = aItems[i+1]
# Compute delta between left and right item
# We remember that (right >= left + 1
nDelta = item2 - (item1 + 1)
if(nDelta < 0):
# Fix the right item
fixes[i+1] -= nDelta
aItems[i+1] -= nDelta
# Need another loop
bNeedFix = True
# Compute the fix size (rounded up)
# max(s) should be int and the division should produce an int
nFix = int((max(fixes)+1)/2)
print("current nFix:",nFix)
# Balance all fixes
for i in range(len(aItems)):
fixes[i] -= nFix
print("final Fixes:",fixes)
print("d:",nFix)
print("original sequence:",seq[1:])
print("result sequence:",aItems)
return
Here's whats displayed:
Working with: [6, 2, 4, 8, 3, 1, 12]
[0]= 6 So the following numbers are the sequence:
aItems = [2, 4, 8, 3, 1, 12]
fixes>> [0, 0, 0, 0, 0, 0]
current nFix: 6
final Fixes: [-6, -6, -6, 0, 3, -6]
d: 1
original sequence: [2, 4, 8, 3, 1, 12]
result sequence: [2, 4, 8, 9, 10, 12]
d SHOULD be: 5
done!
~Note~
I start at 1 rather than 0 due to the first element being a key
As anticipated, here is (or should be) the Python version of my initial solution:
def ComputeMaxDelta(aItems):
# Create a copy to speed up comparison on modified values
aItems = aItems[:]
# Will store the fix values for every item
# this should allocate 'length' times the 0 value
fixes = [0] * len(aItems)
# Loop until no more fixes get applied
bNeedFix = True
while(bNeedFix):
# Hope will have no fix this turn
bNeedFix = False
# loop all subsequent item pairs (i should run from 0 to length - 2)
for i in range(0,len(aItems)-1):
# Left item
item1 = aItems[i]
# right item
item2 = aItems[i+1]
# Compute delta between left and right item
# We remember that (right >= left + 1
nDelta = item2 - (item1 + 1)
if(nDelta < 0):
# Fix the right item
fixes[i+1] -= nDelta
aItems[i+1] -= nDelta
# Need another loop
bNeedFix = True
# Compute the fix size (rounded up)
# max(s) should be int and the division should produce an int
nFix = (max(fixes)+1)/2 # corrected from **(max(s)+1)/2**
# Balance all fixes
for i in range(len(s)):
fixes[i] -= nFix
print("d:",nFix) # corrected from **print("d:",nDelta)**
print("s:",fixes)
return
I took your Python and fixed in order to operate exactly as my C# solution.
I don't know Python, but looking for some reference on the web, I should have found the points where your porting was failing.
If you compare your python version with mine you should find the following differences:
You saved a reference aItems into s and used it as my fixes, but fixes was meant to start as all 0.
You didn't cloned aItems over itself, then every alteration to its items was reflected outside of the method.
Your for loop was starting at index 1, whereas mine started at 0 (the very first element).
After the check for nDelta you subtracted nDelta from both s and aItems, but as I stated at points 1 and 2 they were pointing to the same items.
The ceil instruction was unnedeed because the division between two integers produces an integer, as with C#.
Please remember that I fixed the Python code basing my knowledge only on online documentation, because I don't code in that language, so I'm not 100% sure about some syntax (my main doubt is about the fixes declaration).
Regards,
Daniele.
Here is my solution:
public static int ComputeMaxDelta(int[] aItems, out int[] fixes)
{
// Create a copy to speed up comparison on modified values
aItems = (int[])aItems.Clone();
// Will store the fix values for every item
fixes = new int[aItems.Length];
// Loop until no more fixes get applied
var bNeedFix = true;
while (bNeedFix)
{
// Hope will have no fix this turn
bNeedFix = false;
// loop all subsequent item pairs
for (int ixItem = 0; ixItem < aItems.Length - 1; ixItem++)
{
// Left item
var item1 = aItems[ixItem];
// right item
var item2 = aItems[ixItem + 1];
// Compute delta between left and right item
// We remember that (right >= left + 1)
var nDelta = item2 - (item1 + 1);
if (nDelta < 0)
{
// Fix the right item
fixes[ixItem + 1] -= nDelta;
aItems[ixItem + 1] -= nDelta;
//Need another loop
bNeedFix = true;
}
}
}
// Compute the fix size (rounded up)
var nFix = (fixes.Max() + 1) / 2;
// Balance all fixes
for (int ixItem = 0; ixItem < aItems.Length; ixItem++)
fixes[ixItem] -= nFix;
return nFix;
}
The function returns the maximum computed fix gap.
As a bounus, the parameter fixes will receive the fixes for every item. These are the delta to apply to each source value in order to be sure that they will be in ascending order: some fix can be reduced but some analysis loop is required to achieve that optimization.
The following is a code to test the algorithm. If you set a breakpoint at the end of the loop, you'll be able to check the result for sequence you provided in your example.
var random = new Random((int)Stopwatch.GetTimestamp());
for (int ixLoop = -1; ixLoop < 100; ixLoop++)
{
int nCount;
int[] aItems;
// special case as the provided sample sequence
if (ixLoop == -1)
{
aItems = new[] { 2, 4, 8, 3, 1, 12 };
nCount = aItems.Length;
}
else
{
// Generates a random amount of items based on my screen's width
nCount = 4 + random.Next(21);
aItems = new int[nCount];
for (int ixItem = 0; ixItem < nCount; ixItem++)
{
// Keep the generated numbers below 30 for easier human analysis
aItems[ixItem] = random.Next(30);
}
}
Console.WriteLine("***");
Console.WriteLine(" # " + GetText(Enumerable.Range(0, nCount).ToArray()));
Console.WriteLine(" " + GetText(aItems));
int[] aFixes;
var nFix = ComputeMaxDelta(aItems, out aFixes);
// Computes the new values, that will be always in ascending order
var aNew = new int[aItems.Length];
for (int ixItem = 0; ixItem < aItems.Length; ixItem++)
{
aNew[ixItem] = aItems[ixItem] + aFixes[ixItem];
}
Console.WriteLine(" = " + nFix.ToString());
Console.WriteLine(" ! " + GetText(aFixes));
Console.WriteLine(" > " + GetText(aNew));
}
Regards,
Daniele.
I would like to split a rectangle in cells. In each cell it should be create a random coordinate (y, z).
The wide and height of the rectangle are known (initialW / initalH).
The size of the cells are calculated (dy / dz).
The numbers, in how many cells the rectangle to be part, are known. (numberCellsY / numberCellsZ)
Here my Code in Fortran to split the rectangle in Cells:
yRVEMin = 0.0
yRVEMax = initialW
dy = ( yRVEMax - yRVEMin ) / numberCellsY
zRVEMin = 0.0
zRVEMax = initialH
dz = ( zRVEMax - zRVEMin ) / numberCellsZ
do i = 1, numberCellsY
yMin(i) = (i-1)*dy
yMax(i) = i*dy
end do
do j = 1, numberCellsZ
zMin(j) = (j-1)*dz
zMax(j) = j*dz
end do
Now I would like to produce a random coordinate in each cell. The problem for me is, to store the coodinates in an array. It does not necessarily all be stored in one array, but as least as possible.
To fill the cells with coordinates it should start at the bottom left cell, go through the rows (y-direction), and after the last cell (numberCellsY) jump a column higher (z-dicrection) and start again by the first cell of the new row at left side. That should be made so long until a prescribed number (nfibers) is reached.
Here a deplorable try to do it:
call random_seed
l = 0
do k = 1 , nfibers
if (l < numberCellsY) then
l = l + 1
else
l = 1
end if
call random_number(y)
fiberCoordY(k) = yMin(l) + y * (yMax(l) - yMin(l))
end do
n = 0
do m = 1 , nfibers
if (n < numberCellsZ) then
n = n + 1
else
n = 1
end if
call random_number(z)
fiberCoordZ(m) = zMin(n) + z * (zMax(n) - zMin(n))
end do
The output is not what I want! fiberCoordZ should be stay on (zMin(1) / zMax(1) as long as numberCellsY-steps are reached.
The output for following settings:
nfibers = 9
numberCellsY = 3
numberCellsZ = 3
initialW = 9.0
initialH = 9.0
My random output for fiberCoordY is:
1.768946 3.362770 8.667685 1.898700 5.796713 8.770239 2.463412 3.546694 7.074708
and for fiberCoordZ is:
2.234807 5.213032 6.762228 2.948657 5.937295 8.649946 0.6795220 4.340364 8.352566
In this case the first 3 numbers of fiberCoordz should have a value between 0.0 and 3.0. Than number 4 - 6 a value between 3.0 and 6.0. And number 7 - 9 a value bewtween 6.0 - 9.0.
How can I solve this? If somebody has a solution with a better approach, please post it!
Thanks
Looking at
n = 0
do m = 1 , nfibers
if (n < numberCellsZ) then
n = n + 1
else
n = 1
end if
call random_number(z)
fiberCoordZ(m) = zMin(n) + z * (zMax(n) - zMin(n))
end do
we see that the z coordinate offset (the bottom cell boundary of interest) is being incremented inappropriately: for each consecutive nfibers/numberCellsZ coordinates n should be constant.
n should be incremented only every numberCellsY iterations, so perhaps a condition like
if (MOD(m, numberCellsY).eq.1) n=n+1
would be better.
Thanks francescalus! It works fine.
I added a little more for the case that nfibers > numberCellsY*numberCellsZ
n=0
do m = 1 , nfibers
if (MOD(m, numberCellsY).eq.1 .and. (n < numberCellsY)) then
n=n+1
end if
if (MOD(m, numberCellsY*numberCellsZ).eq.1 ) then
n = 1
end if
call random_number(z)
fiberCoordZ(m) = zMin(n) + z * (zMax(n) - zMin(n))
end do
I am using Excel 2007 which supports Columns upto 16,384 Columns. I would like to obtain the Column name corresponding Column Number.
Currently, I am using the following code. However this code supports upto 256 Columns. Any idea how to obtain Column Name if the column number is greater than 256.
function loc = xlcolumn(column)
if isnumeric(column)
if column>256
error('Excel is limited to 256 columns! Enter an integer number <256');
end
letters = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
count = 0;
if column-26<=0
loc = char(letters(column));
else
while column-26>0
count = count + 1;
column = column - 26;
end
loc = [char(letters(count)) char(letters(column))];
end
else
letters = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
if size(column,2)==1
loc =findstr(column,letters);
elseif size(column,2)==2
loc1 =findstr(column(1),letters);
loc2 =findstr(column(2),letters);
loc = (26 + 26*loc1)-(26-loc2);
end
end
Thanks
As a diversion, here is an all function handle example, with (almost) no file-based functions required. This is based on the dec2base function, since Excel column names are (almost) base 26 numbers, with the frustrating difference that there are no "0" characters.
Note: this is probably a terrible idea overall, but it works. Better solutions are probably found elsewhere in the file exchange.
First, the one file based function that I couldn't get around, to perform arbitrary depth function composition.
function result = compose( fnHandles )
%COMPOSE Compose a set of functions
% COMPOSE({fnHandles}) returns a function handle consisting of the
% composition of the cell array of input function handles.
%
% For example, if F, G, and H are function handles with one input and
% one output, then:
% FNCOMPOSED = COMPOSE({F,G,H});
% y = FNCOMPOSED(x);
% is equivalent to
% y = F(G(H(x)));
if isempty(fnHandles)
result = #(x)x;
elseif length(fnHandles)==1
result = fnHandles{1};
else
fnOuter = fnHandles{1};
fnRemainder = compose(fnHandles(2:end));
result = #(x)fnOuter(fnRemainder(x));
end
Then, the bizarre, contrived path to convert base26 values into the correct string
%Functions leading to "getNumeric", which creates a numeric, base26 array
remapUpper = #(rawBase)(rawBase + (rawBase>='A')*(-55)); %Map the letters 'A-P' to [10:26]
reMapLower = #(rawBase)(rawBase + (rawBase<'A')*(-48)); %Map characters '0123456789' to [0:9]
getRawBase = #(x)dec2base(x, 26);
getNumeric = #(x)remapUpper(reMapLower(getRawBase(x)));
%Functions leading to "correctNumeric"
% This replaces zeros with 26, and reduces the high values entry by 1.
% Similar to "borrowing" as we learned in longhand subtraction
borrowDownFrom = #(x, fromIndex) [x(1:(fromIndex-1)) (x(fromIndex)-1) (x(fromIndex+1)+26) (x((fromIndex+2):end))];
borrowToIfNeeded = #(x, toIndex) (x(toIndex)<=0)*borrowDownFrom(x,toIndex-1) + (x(toIndex)>0)*(x); %Ugly numeric switch
getAllConditionalBorrowFunctions = #(numeric)arrayfun(#(index)#(numeric)borrowToIfNeeded(numeric, index),(2:length(numeric)),'uniformoutput',false);
getComposedBorrowFunction = #(x)compose(getAllConditionalBorrowFunctions(x));
correctNumeric = #(x)feval(getComposedBorrowFunction(x),x);
%Function to replace numerics with letters, and remove leading '#' (leading
%zeros)
numeric2alpha = #(x)regexprep(char(x+'A'-1),'^#','');
%Compose complete function
num2ExcelName = #(x)arrayfun(#(x)numeric2alpha(correctNumeric(getNumeric(x))), x, 'uniformoutput',false)';
Now test using some stressing transitions:
>> num2ExcelName([1:5 23:28 700:704 727:729 1024:1026 1351:1355 16382:16384])
ans =
'A'
'B'
'C'
'D'
'E'
'W'
'X'
'Y'
'Z'
'AA'
'AB'
'ZX'
'ZY'
'ZZ'
'AAA'
'AAB'
'AAY'
'AAZ'
'ABA'
'AMJ'
'AMK'
'AML'
'AYY'
'AYZ'
'AZA'
'AZB'
'AZC'
'XFB'
'XFC'
'XFD'
This function I wrote works for any number of columns (until Excel runs out of columns). It just requires a column number input (e.g. 16368 will return a string 'XEN').
If the application of this concept is different than my function, it's important to note that a column of x number of A's begins every 26^(x-1) + 26^(x-2) + ... + 26^2 + 26 + 1. (e.g. 'AAA' begins on 26^2 + 26 + 1 = 703)
function [col_str] = let_loc(num_loc)
test = 2;
old = 0;
x = 0;
while test >= 1
old = 26^x + old;
test = num_loc/old;
x = x + 1;
end
num_letters = x - 1;
str_array = zeros(1,num_letters);
for i = 1:num_letters
loc = floor(num_loc/(26^(num_letters-i)));
num_loc = num_loc - (loc*26^(num_letters-i));
str_array(i) = char(65 + (loc - 1));
end
col_str = strcat(str_array(1:length(str_array)));
end
Hope this saves someone some time!