[1..4] could help to generate
[1;2;3;4]
But I wish to generate a range like this:
[10;8;6;4;2]
How to use range semantics to achieve this(interval+descending)? Is a "for" loop mandatory in this case?
Thanks a lot.
Try this:
[10 .. -2 .. 2] ;;
// val it : int list = [10; 8; 6; 4; 2]
The value in the middle specifies the interval.
Related
If I have two unknown values, lets say x and y, what is the best way loop through all of the values between between those values?
For example, given the values x = 0 and y = 5 I would like to do something with the values 0, 1, 2, 3, 4, and 5. The result could exclude 0 and 5 if this is simpler.
Using Swift's Range operator, I could do something like this:
for i in x...y {
// Do something with i
}
Except I do not know if x or y is the greater value.
The Swift documentation for Range Operators states:
The closed range operator (a...b) defines a range that runs from a to b, and includes the values a and b. The value of a must not be greater than b.
There are a number of solutions here. A pretty straight forward one is:
let diff = y - x
for i in 0...abs(diff) {
let value = min(x, y) + i
// Do something with value
}
Is there a better, or more elegant way to achieve this?
I guess the most explicit way of writing it would be:
for i in min(a, b)...max(a, b) {
// Do something with i
}
To exclude the first and last value, you can increment your lower limit and use the Swift ..< syntax:
let lowerLimit = min(a, b) + 1
let upperLimit = max(a, b)
for i in lowerLimit..<upperLimit {
// Do something with i
}
Hello experienced pythoners.
The goal is simply to read in my own files which have the following format, and to then apply mathematical operations to these values and polynomials. The files have the following format:
m1:=10:
m2:=30:
Z1:=1:
Z2:=-1:
...
Some very similar variables, next come the laguerre polynomials
...
F:= (12.58295)*L(0,x)*L(1,y)*L(6,z) + (30.19372)*L(0,x)*L(2,y)*L(2,z) - ...:
Where L stands for a laguerre polynomial and takes two arguments.
I have written a procedure in Python which splits apart each line into a left and right hand side split using the "=" character as a divider. The format of these files is always the same, but the number of laguerre polynomials in F can vary.
import re
linestring = open("file.txt", "r").read()
linestring = re.sub("\n\n","\n",str(linestring))
linestring = re.sub(",\n",",",linestring)
linestring = re.sub("\\+\n","+",linestring)
linestring = re.sub(":=\n",":=",linestring)
linestring = re.sub(":\n","\n",linestring)
linestring = re.sub(":","",linestring)
LINES = linestring.split("\n")
for LINE in LINES:
LINE = re.sub(" ","",LINE)
print "LINE=", LINE
if len(LINE) <=0:
next
PAIR = LINE.split("=")
print "PAIR=", PAIR
LHS = PAIR[0]
RHS = PAIR[1]
print "LHS=", LHS
print "RHS=", RHS
The first re.sub block just deals with formatting the file and discarding characters that python will not be able to process; then a loop is performed to print 4 things, LINE, PAIR, LHS and RHS, and it does this nicely. using the example file from above the procedure will print the following:
LINE= m1=1
PAIR= ['m1', '1']
LHS= m1
RHS= 1
LINE= m2=1
PAIR= ['m2', '1']
LHS= m2
RHS= 1
LINE= Z1=-1
PAIR= ['Z1', '-1']
LHS= Z1
RHS= -1
LINE= Z2=-1
PAIR= ['Z2', '-1']
LHS= Z2
RHS= -1
LINE= F= 12.5*L(0,x)L(1,y) + 30*L(0,x)L(2,y)L(2,z)
PAIR=['F', '12.5*L(0,x)L(1,y) + 30*L(0,x)L(2,y)L(2,z)']
LHS= F
RHS= 12.5*L(0,x)L(1,y) + 30*L(0,x)L(2,y)L(2,z)
My question is what is the next best step to process this output and use it in a mathematical script, especially assigning the L to mean a laguerre polynomial? I tried putting the LHS and RHS into a dictionary, but found it troublesome to put F in it due to the laguerre polynomials.
Any ideas are welcome. Perhaps I am overcomplicating this and there is a much simpler way to parse this file.
Many thanks in advance
Your parsing algorithm doesn't seem to work correctly, as the RHS of your variables dont produce the expected result.
Also the first re.sub block where you want to format the file seems overly complicated. Assuming every statement in your input file is terminated by a colon, you could get rid of all whitespace and newlines and seperate the statements using the following code:
linestring = open('file.txt','r').read()
strippedstring = linestring.replace('\n','').replace(' ','')
statements = re.split(':(?!=)',strippedstring)[:-1]
Then you iterate over the statements and split each one in LHS and RHS:
for st in statements:
lhs,rhs = re.split(':=',st)
print 'lhs=',lhs
print 'rhs=',rhs
In the next step, try to distinguish normal float variables and polynomials:
#evaluate rhs
try:
#interpret as numeric constant
f = float(rhs)
print " ",f
except ValueError:
#interpret as laguerre-polynomial
summands = re.split('\+', re.sub('-','+-',rhs))
for s in summands:
m = re.match("^(?P<factor>-?[0-9]*(\.[0-9]*)?)(?P<poly>(\*?L\([0-9]+,[a-z]\))*)", s)
if not m:
print ' polynomial misformatted'
continue
f = m.group('factor')
print ' factor: ',f
p = m.group('poly')
for l in re.finditer("L\((?P<a>[0-9]+),(?P<b>[a-z])\)",p):
print ' poly: L(%s,%s)' % (l.group("a"),l.group("b"))
This should work for your given example file.
I am playing a little bit with Lua.
I came across the following code snippet that have an unexpected behavior:
a = 3;
b = 5;
c = a-- * b++; // some computation
print(a, b, c);
Lua runs the program without any error but does not print 2 6 15 as expected. Why ?
-- starts a single line comment, like # or // in other languages.
So it's equivalent to:
a = 3;
b = 5;
c = a
LUA doesn't increment and decrement with ++ and --. -- will instead start a comment.
There isn't and -- and ++ in lua.
so you have to use a = a + 1 or a = a -1 or something like that
If you want 2 6 15 as the output, try this code:
a = 3
b = 5
c = a * b
a = a - 1
b = b + 1
print(a, b, c)
This will give
3 5 3
because the 3rd line will be evaluated as c = a.
Why? Because in Lua, comments starts with --. Therefore, c = a-- * b++; // some computation is evaluated as two parts:
expression: c = a
comment: * b++; //// some computation
There are 2 problems in your Lua code:
a = 3;
b = 5;
c = a-- * b++; // some computation
print(a, b, c);
One, Lua does not currently support incrementation. A way to do this is:
c = a - 1 * b + 1
print(a, b, c)
Two, -- in Lua is a comment, so using a-- just translates to a, and the comment is * b++; // some computation.
Three, // does not work in Lua, use -- for comments.
Also it's optional to use ; at the end of every line.
You can do the following:
local default = 0
local max = 100
while default < max do
default = default + 1
print(default)
end
EDIT: Using SharpLua in C# incrementing/decrementing in lua can be done in shorthand like so:
a+=1 --increment by some value
a-=1 --decrement by some value
In addition, multiplication/division can be done like so:
a*=2 --multiply by some value
a/=2 --divide by some value
The same method can be used if adding, subtracting, multiplying or dividing one variable by another, like so:
a+=b
a-=b
a/=b
a*=b
This is much simpler and tidier and I think a lot less complicated, but not everybody will share my view.
Hope this helps!
I have written the following code to generate all possible combinations of some numbers:
let allCombinations (counts:int[]) =
let currentPositions = Array.create (counts.Length) 0
let idx = ref (counts.Length-1)
seq{
while currentPositions.[0]<counts.[0] do
yield currentPositions
currentPositions.[!idx]<-currentPositions.[!idx]+1
while currentPositions.[!idx] >= counts.[!idx] && !idx>=1 do
currentPositions.[!idx]<-0
idx:=!idx-1
currentPositions.[!idx]<-currentPositions.[!idx]+1
idx:=counts.Length-1
}
I am consuming the sequence in some other part of the program like this:
allCombinations counts |> Seq.map (fun idx -> buildGuess n digitsPerPos idx) ...
So far so good. The programs runs as expected and generates the combinations. For input [|2;2;2|] it generates the eight values:
[|0; 0; 0|]
[|0; 0; 1|]
[|0; 1; 0|]
[|0; 1; 1|]
[|1; 0; 0|]
[|1; 0; 1|]
[|1; 1; 0|]
[|1; 1; 1|]
However when I use PSeq to parallelise the generated sequence all values to be consumed change to [|2;0;0|] which is the last value of the currentPositions array in the while loop above.
If i use
yield (currentPositions|>Array.copy)
instead of
yield currentPositions
everything works ok in both sequential and parallel versions.
Why does this happen; Is there a most efficient way to yield the result; Thank you in advance;
The problem is that you're creating a single array which you're mutating between iterations.
You can take the parallelism out of the equation by building a list of the results instead of printing them out one by one - if you build the list first and then print them all, you'll see the same result; the list will contain the same reference 8 times, always to the same array instance.
Basically to avoid side-effects, you need each result to be independent of the other - so you should create a separate array each time.
Consider the following code:
let dl = 9.5 / 11.
let min = 21.5 + dl
let max = 40.5 - dl
let a = [ for z in min .. dl .. max -> z ] // should have 21 elements
let b = a.Length
"a" should have 21 elements but has got only 20 elements. The "max - dl" value is missing. I understand that float numbers are not precise, but I hoped that F# could work with that. If not then why F# supports List comprehensions with float iterator? To me, it is a source of bugs.
Online trial: http://tryfs.net/snippets/snippet-3H
Converting to decimals and looking at the numbers, it seems the 21st item would 'overshoot' max:
let dl = 9.5m / 11.m
let min = 21.5m + dl
let max = 40.5m - dl
let a = [ for z in min .. dl .. max -> z ] // should have 21 elements
let b = a.Length
let lastelement = List.nth a 19
let onemore = lastelement + dl
let overshoot = onemore - max
That is probably due to lack of precision in let dl = 9.5m / 11.m?
To get rid of this compounding error, you'll have to use another number system, i.e. Rational. F# Powerpack comes with a BigRational class that can be used like so:
let dl = 95N / 110N
let min = 215N / 10N + dl
let max = 405N / 10N - dl
let a = [ for z in min .. dl .. max -> z ] // Has 21 elements
let b = a.Length
Properly handling float precision issues can be tricky. You should not rely on float equality (that's what list comprehension implicitely does for the last element). List comprehensions on float are useful when you generate an infinite stream. In other cases, you should pay attention to the last comparison.
If you want a fixed number of elements, and include both lower and upper endpoints, I suggest you write this kind of function:
let range from to_ count =
assert (count > 1)
let count = count - 1
[ for i = 0 to count do yield from + float i * (to_ - from) / float count]
range 21.5 40.5 21
When I know the last element should be included, I sometimes do:
let a = [ for z in min .. dl .. max + dl*0.5 -> z ]
I suspect the problem is with the precision of floating point values. F# adds dl to the current value each time and checks if current <= max. Because of precision problems, it might jump over max and then check if max+ε <= max (which will yield false). And so the result will have only 20 items, and not 21.
After running your code, if you do:
> compare a.[19] max;;
val it : int = -1
It means max is greater than a.[19]
If we do calculations the same way the range operator does but grouping in two different ways and then compare them:
> compare (21.5+dl+dl+dl+dl+dl+dl+dl+dl) ((21.5+dl)+(dl+dl+dl+dl+dl+dl+dl));;
val it : int = 0
> compare (21.5+dl+dl+dl+dl+dl+dl+dl+dl+dl) ((21.5+dl)+(dl+dl+dl+dl+dl+dl+dl+dl));;
val it : int = -1
In this sample you can see how adding 7 times the same value in different order results in exactly the same value but if we try it 8 times the result changes depending on the grouping.
You're doing it 20 times.
So if you use the range operator with floats you should be aware of the precision problem.
But the same applies to any other calculation with floats.