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I'm having problem returning spesific amount of decimal numbers from this function, i would like it to get that info from "dec" argument, but i'm stuck with this right now.
Edit: Made it work with the edited version bellow but isn't there a better way?
local function remove_decimal(t, dec)
if type(dec) == "number" then
for key, num in pairs(type(t) == "table" and t or {}) do
if type(num) == "number" then
local num_to_string = tostring(num)
local mod, d = math.modf(num)
-- find only decimal numbers
local num_dec = num_to_string:sub(#tostring(mod) + (mod == 0 and num < 0 and 3 or 2))
if dec <= #num_dec then
-- return amount of deciamls in the num by dec
local r = d < 0 and "-0." or "0."
local r2 = r .. num_dec:sub(1, dec)
t[key] = mod + tonumber(r2)
end
end
end
end
return t
end
By passing the function bellow i want a result like this:
result[1] > 0.12
result[2] > -0.12
result[3] > 123.45
result[4] > -1.23
local result = remove_decimal({0.123, -0.123, 123.456, -1.234}, 2)
print(result[1])
print(result[2])
print(result[3])
print(result[4])
I tried this but it seems to only work with one integer numbers and if number is 12.34 instead of 1.34 e.g, the decimal place will be removed and become 12.3. Using other methods
local d = dec + (num < 0 and 2 or 1)
local r = tonumber(num_to_string:sub(1, -#num_to_string - d)) or 0
A good approach is to find the position of the decimal point (the dot, .) and then extract a substring starting from the first character to the dot's position plus how many digits you want:
local function truncate(number, dec)
local strnum = tostring(number)
local i, j = string.find(strnum, '%.')
if not i then
return number
end
local strtrn = string.sub(strnum, 1, i+dec)
return tonumber(strtrn)
end
Call it like this:
print(truncate(123.456, 2))
print(truncate(1234567, 2))
123.45
1234567
To bulk-truncate a set of numbers:
local function truncate_all(t, dec)
for key, value in pairs(t) do
t[key] = truncate(t[key], dec)
end
return t
end
Usage:
local result = truncate_all({0.123, -0.123, 123.456, -1.234}, 2)
for key, value in pairs(result) do
print(key, value)
end
1 0.12
2 -0.12
3 123.45
4 -1.23
One could use the function string.format which is similar to the printf functions from C language. If one use the format "%.2f" the resulting string will contain 2 decimals, if one use "%.3f" the resulting string will be contain 3 decimals, etc. The idea is to dynamically create the format "%.XXXf" corresponding to the number of decimal needed by the function. Then call the function string.format with the newly created format string to generate the string "123.XXX". The last step would be to convert back the string to a number with the function tonumber.
Note that if one want the special character % to be preserved when string.format is called, you need to write %%.
function KeepDecimals (Number, DecimalCount)
local FloatFormat = string.format("%%.%df", DecimalCount)
local String = string.format(FloatFormat, Number)
return tonumber(String)
end
The behavior seems close to what the OP is looking for:
for Count = 1, 5 do
print(KeepDecimals(1.123456789, Count))
end
This code should print the following:
1.1
1.12
1.123
1.1235
1.12346
Regarding the initial code, it's quite straight-forward to integrate the provided solution. Note that I renamed the function to keep_decimal because in my understanding, the function will keep the requested number of decimals, and discard the rest.
function keep_decimal (Table, Count)
local NewTable = {}
local NewIndex = 1
for Index = 1, #Table do
NewTable[NewIndex] = KeepDecimal(Table[Index], Count)
NewIndex = NewIndex + 1
end
return NewTable
end
Obviously, the code could be tested easily, simply by copy and pasting into a Lua interpreter.
Result = keep_decimal({0.123, -0.123, 123.456, -1.234}, 2)
for Index = 1, #Result do
print(Result[Index])
end
This should print the following:
0.12
-0.12
123.46
-1.23
Edit due to the clarification of the need of truncate:
function Truncate (Number, Digits)
local Divider = Digits * 10
local TruncatedValue = math.floor(Number * Divider) / Divider
return TruncatedValue
end
On my computer, the code is working as expected:
> Truncate(123.456, 2)
123.45
I am new to programming in LUA. And I am not able to solve this question below.
Given a number N, generate a star pattern such that on the first line there are N stars and on the subsequent lines the number of stars decreases by 1.
The pattern generated should have N rows. In every row, every fifth star (*) is replaced with a hash (#). Every row should have the required number of stars (*) and hash (#) symbols.
Sample input and output, where the first line is the number of test cases
This is what I tried.. And I am not able to move further
function generatePattern()
n = tonumber(io.read())
i = n
while(i >= 1)
do
j = 1
while(j<=i)
do
if(j<=i)
then
if(j%5 == 0)
then
print("#");
else
print("*");
end
print(" ");
end
j = j+1;
end
print("\n");
i = i-1;
end
end
tc = tonumber(io.read())
for i=1,tc
do
generatePattern()
end
First, just the stars without hashes. This part is easy:
local function pattern(n)
for i=n,1,-1 do
print(string.rep("*", i))
end
end
To replace each 5th asterisk with a hash, you can extend the expression with the following substitution:
local function pattern(n)
for i=n,1,-1 do
print((string.rep("*", i):gsub("(%*%*%*%*)%*", "%1#")))
end
end
The asterisks in the pattern need to be escaped with a %, since * holds special meaning within Lua patterns.
Note that string.gsub returns 2 values, but they can be truncated to one value by adding an extra set of parentheses, leading to the somewhat awkward-looking form print((..)).
Depending on Lua version the metamethod __index holding rep for repeats...
--- Lua 5.3
n=10
asterisk='*'
print(asterisk:rep(n))
-- puts out: **********
#! /usr/bin/env lua
for n = arg[1], 1, -1 do
local char = ''
while #char < n do
if #char %5 == 4 then char = char ..'#'
else char = char ..'*'
end -- mod 5
end -- #char
print( char )
end -- arg[1]
chmod +x asterisk.lua
./asterisk.lua 15
Please do not follow this answer since it is bad coding style! I would delete it but SO won't let me. See comment and other answers for better solutions.
My Lua print adds newlines to each printout, therefore I concatenate each character in a string and print the concatenated string out afterwards.
function generatePattern()
n = tonumber(io.read())
i = n
while(i >= 1)
do
ouput = ""
j = 1
while(j<=i)
do
if(j%5 == 0)
then
ouput=ouput .. "#";
else
ouput=ouput .. "*";
end
j = j+1;
end
print(ouput);
i = i-1;
end
end
Also this code is just yours minimal transformed to give the correct output. There are plenty of different ways to solve the task, some are faster or more intuitive than others.
I have a macro that searches but does not find the “7” (If Right(pair, 2) = 7 Then). The thing is when I change the number to 11 or 12 etc. (any two digits) and the findXX in the code, it works fine. Does anyone know what’s occurring and what is the exact change I need to do.
Option Explicit
Sub DivideSomeStuff()
Dim pair As Range, accumulator As Range
Dim findSeven As Double
Dim remainder As Long
For Each pair In Range("B30, F30, J30")
If Right(pair, 2) = 7 Then
If pair.Offset(0, 2) <= 12 Then
remainder = 0
Else
remainder = pair.Offset(0, 2) Mod 10
End If
findSeven = (pair.Offset(0, 2) - remainder) / 10
For Each accumulator In Range("A36, D36, G36, J36, M36, A40, D40, G40, J40, M40")
If accumulator.Offset(-1, 0) = Val(Left(pair, InStr(pair, "-") - 1)) Then
accumulator.Value = accumulator.Value + remainder
End If
accumulator.Value = accumulator.Value + findSeven
Next accumulator
End If
Next pair
End Sub
Change it from ...
Right(pair, 2) = 7
... to ...
Right(pair, 1) = 7
You’re currently getting the 2 right values when 7 is a single character.
You may need to put quotes around the 7 too, see if works without them though.
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 writing code in gfortran to separate a variable time stamp into its separate parts of year, month, and day. I have written this code so the user can input what the time stamp format will be (ie. YEAR/MON/DAY, DAY/MON/YEAR, etc). This creates a total of 6 possible combinations. I have written code that attempts to deal with this, but I believe it to be ugly and poorly done.
My current code uses a slew of 'if' and 'goto' statements. The user provides 'tsfo', the time stamp format. 'ts' is a character array containing the time stamp data (as many as 100,000 time stamps). 'tsdelim' is the delimiter between the year, month, and day. I must loop from 'frd' (the first time stamp) to 'nlines' (the last time stamp).
Here is the relevant code.
* Choose which case to go to.
first = INDEX(tsfo,tsdelim)
second = INDEX(tsfo(first+1:),tsdelim) + first
if (INDEX(tsfo(1:first-1),'YYYY') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'MM') .ne. 0) THEN
goto 1001
else
goto 1002
end if
else if (INDEX(tsfo(1:first-1),'MM') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'DD') .ne. 0) THEN
goto 1003
else
goto 1004
end if
else if (INDEX(tsfo(1:first-1),'DD') .ne. 0) THEN
if (INDEX(tsfo(first+1:second-1),'MM') .ne. 0) THEN
goto 1005
else
goto 1006
end if
end if
first = 0
second = 0
* Obtain the Julian Day number of each data entry.
* Acquire the year, month, and day of the time stamp.
* Find 'first' and 'second' and act accordingly.
* Case 1: YYYY/MM/DD
1001 do i = frd,nlines
first = INDEX(ts(i),tsdelim)
second = INDEX(ts(i)(first+1:),tsdelim) + first
read (ts(i)(1:first-1), '(i4)') Y
read (ts(i)(first+1:second-1), '(i2)') M
read (ts(i)(second+1:second+2), '(i2)') D
* Calculate the Julian Day number using a function.
temp1(i) = JLDYNUM(Y,M,D)
end do
goto 1200
* Case 2: YYYY/DD/MM
1002 do i = frd,nlines
first = INDEX(ts(i),tsdelim)
second = INDEX(ts(i)(first+1:),tsdelim) + first
read (ts(i)(1:first-1), '(i4)') Y
read (ts(i)(second+1:second+2), '(i2)') M
read (ts(i)(first+1:second-1), '(i2)') D
* Calculate the Julian Day number using a function.
temp1(i) = JLDYNUM(Y,M,D)
end do
goto 1200
* Onto the next part of the code
1200 blah blah blah
I believe this code will work, but I do not think it is a very good method. Is there a better way to go about this?
It is important to note that the indices 'first' and 'second' must be calculated for each time stamp as the month and day can both be represented by 1 or 2 integers. The year is always represented by 4.
With only six permutations to handle I would just build a look-up table with the whole tsfo string as the key and the positions of year, month and day (1st, 2nd or 3rd) as the values. Any unsupported formats should produce an error, which I haven't coded below. When subsequently you loop though your ts list and split an item you know which positions to cast to the year, month and day integer variables:
PROGRAM timestamp
IMPLICIT NONE
CHARACTER(len=10) :: ts1(3) = ["2000/3/4 ","2000/25/12","2000/31/07"]
CHARACTER(len=10) :: ts2(3) = ["3/4/2000 ","25/12/2000","31/07/2000"]
CALL parse("YYYY/DD/MM",ts1)
print*
CALL parse("DD/MM/YYYY",ts2)
CONTAINS
SUBROUTINE parse(tsfo,ts)
IMPLICIT NONE
CHARACTER(len=*),INTENT(in) :: tsfo, ts(:)
TYPE sti
CHARACTER(len=10) :: stamp = "1234567890"
INTEGER :: iy = -1, im = -1, id = -1
END TYPE sti
TYPE(sti),PARAMETER :: stamps(6) = [sti("YYYY/MM/DD",1,2,3), sti("YYYY/DD/MM",1,3,2),&
sti("MM/DD/YYYY",2,3,1), sti("DD/MM/YYYY",3,2,1),&
sti("MM/YYYY/DD",2,1,3), sti("DD/YYYY/MM",3,1,2)]
TYPE(sti) :: thisTsfo
INTEGER :: k, k1, k2
INTEGER :: y, m, d
CHARACTER(len=10) :: cc(3)
DO k=1,SIZE(stamps)
IF(TRIM(tsfo) == stamps(k)%stamp) THEN
thisTsfo = stamps(k)
EXIT
ENDIF
ENDDO
print*,thisTsfo
DO k=1,SIZE(ts)
k1 = INDEX(ts(k),"/")
k2 = INDEX(ts(k),"/",BACK=.TRUE.)
cc(1) = ts(k)(:k1-1)
cc(2) = ts(k)(k1+1:k2-1)
cc(3) = ts(k)(k2+1:)
READ(cc(thisTsfo%iy),'(i4)') y
READ(cc(thisTsfo%im),'(i2)') m
READ(cc(thisTsfo%id),'(i2)') d
PRINT*,ts(k),y,m,d
ENDDO
END SUBROUTINE parse
END PROGRAM timestamp
I would encode the different cases in another way, like this:
module foo
implicit none
private
public encode_datecode
contains
integer function encode_datecode(datestr, sep)
character(len=*), intent(in) :: datestr, sep
integer :: first, second
character(len=1) :: c1, c2, c3
first = index(datestr, sep)
second = index(datestr(first+1:), sep) + first
c1 = datestr(1:1)
c2 = datestr(first+1:first+1)
c3 = datestr(second+1:second+1)
foo = num(c1) + 3*num(c2) + 9*num(c3)
end function encode_datecode
integer function num(c)
character(len=1) :: c
if (c == 'Y') then
num = 0
else if (c == 'M') then
num = 1
else if (c == 'D') then
num = 2
else
stop "Illegal character"
end if
end function num
end module foo
and then handle the legal cases (21, 15, 19, 7, 11, 5) in a SELECT statement.
This takes advantage of the fact that there won't be a 'YDDY/MY/YM' format.
If you prefer better binary or decimal readability, you can also multiply by four or by 10 instead of 3.