I'd like to format a number to look like 1,234 or 1,234,432 or 123,456,789, you get the idea. I tried doing this as follows:
function reformatint(i)
local length = string.len(i)
for v = 1, math.floor(length/3) do
for k = 1, 3 do
newint = string.sub(mystring, -k*v)
end
newint = ','..newint
end
return newint
end
As you can see, a failed attempt, my problem is that I can't figure out what the error is because the program I am running this in refuses to report an error back to me.
Here's a function that takes negative numbers, and fractional parts into account:
function format_int(number)
local i, j, minus, int, fraction = tostring(number):find('([-]?)(%d+)([.]?%d*)')
-- reverse the int-string and append a comma to all blocks of 3 digits
int = int:reverse():gsub("(%d%d%d)", "%1,")
-- reverse the int-string back remove an optional comma and put the
-- optional minus and fractional part back
return minus .. int:reverse():gsub("^,", "") .. fraction
end
assert(format_int(1234) == '1,234')
assert(format_int(1234567) == '1,234,567')
assert(format_int(123456789) == '123,456,789')
assert(format_int(123456789.1234) == '123,456,789.1234')
assert(format_int(-123456789.) == '-123,456,789')
assert(format_int(-123456789.1234) == '-123,456,789.1234')
assert(format_int('-123456789.1234') == '-123,456,789.1234')
print('All tests passed!')
Well, let's take this from the top down. First of all, it's failing because you've got a reference error:
...
for k = 1, 3 do
newint = string.sub(mystring, -k*v) -- What is 'mystring'?
end
...
Most likely you want i to be there, not mystring.
Second, while replacing mystring with i will fix the errors, it still won't work correctly.
> =reformatint(100)
,100
> =reformatint(1)
,000
That's obviously not right. It seems like what you're trying to do is go through the string, and build up the new string with the commas added. But there are a couple of problems...
function reformatint(i)
local length = string.len(i)
for v = 1, math.floor(length/3) do
for k = 1, 3 do -- What is this inner loop for?
newint = string.sub(mystring, -k*v) -- This chops off the end of
-- your string only
end
newint = ','..newint -- This will make your result have a ',' at
-- the beginning, no matter what
end
return newint
end
With some rework, you can get a function that work.
function reformatint(integer)
for i = 1, math.floor((string.len(integer)-1) / 3) do
integer = string.sub(integer, 1, -3*i-i) ..
',' ..
string.sub(integer, -3*i-i+1)
end
return integer
end
The function above seems to work correctly. However, it's fairly convoluted... Might want to make it more readable.
As a side note, a quick google search finds a function that has already been made for this:
function comma_value(amount)
local formatted = amount
while true do
formatted, k = string.gsub(formatted, "^(-?%d+)(%d%d%d)", '%1,%2')
if (k==0) then
break
end
end
return formatted
end
You can do without loops:
function numWithCommas(n)
return tostring(math.floor(n)):reverse():gsub("(%d%d%d)","%1,")
:gsub(",(%-?)$","%1"):reverse()
end
assert(numWithCommas(100000) == "100,000")
assert(numWithCommas(100) == "100")
assert(numWithCommas(-100000) == "-100,000")
assert(numWithCommas(10000000) == "10,000,000")
assert(numWithCommas(10000000.00) == "10,000,000")
The second gsub is needed to avoid -,100 being generated.
I remember discussing about this in the LÖVE forums ... let me look for it...
Found it!
This will work with positive integers:
function reformatInt(i)
return tostring(i):reverse():gsub("%d%d%d", "%1,"):reverse():gsub("^,", "")
end
On the link above you may read details about implementation.
Related
I have created a function that (pseudo)randomly creates a table containing numbers. I then loop this function until at least correct result is found. As soon as I've confirmed that at least one such result exists, I stop the function and return the table.
When I create tables containing small values, there are no issues. However, once the random numbers grow to the range of hundreds, the function begins to return nil, even though the table is true the line before I return it.
local sort = table.sort
local random = math.random
local aMin, aMax = 8, 12
local bMin, bMax = 200, 2000
local function compare( a, b )
return a < b
end
local function getNumbers()
local valid = false
local numbers = {}
-- Generate a random length table, containing random number values.
for i = 1, random( aMin, aMax ) do
numbers[i] = random( bMin, bMax )
end
sort( numbers, compare )
-- See if a specific sequence of numbers exist in the table.
for i = 2, #numbers do
if numbers[i-1]+1 == numbers[i] or numbers[i-1] == numbers[i] then
-- Sequence found, so stop.
valid = true
break
end
end
for i = 1, #numbers-1 do
for j = i+1, #numbers do
if numbers[j] % numbers[i] == 0 and numbers[i] ~= 1 then
valid = true
break
end
end
end
if valid then
print( "Within function:", numbers )
return numbers
else
getNumbers()
end
end
local numbers = getNumbers()
print( "Outside function:", numbers )
This function, to my understanding, is supposed to loop infinitely until I find a valid sequence. The only way that the function can even end, according to my code, is if valid is true.
Sometimes, more often than not, with large numbers the function simply outputs a nil value to the outside of the function. What is going on here?
You're just doing getNumbers() to recurse instead of return getNumbers(). This means that if the recursion gets entered, the final returned value will be nil no matter what else happens.
In the else case of the if valid then, you are not returning anything. You only return anything in the valid case. In the else case, a recursive call may return something, but then you ignore that returned value. The print you see is corresponding to the return from the recursive call; it isn't making it out the original call.
You mean to return getNumbers().
I don't understand why the following code produces error.
The code begins with the main() function at the bottom.
heads = {}
function push(t)
if (#t == 2) then
table.insert(heads, t)
end
end
function remove(id)
for i = 1, #heads do
if (heads[i][2] == id) then
table.remove(heads, i)
end
end
end
function main()
push({50, 1})
push({50, 2})
push({50, 3})
remove(2)
end
When I run the code, I get attempt to index a nil value (field '?') error.
I expect to push the subtable elements into the table and then remove only the second one. So the resulting elements can be {50, 1} and {50, 3}.
Why is my code not working and how to fix this?
Andrew got it right. Never try to remove a value inside a table when you iterate the table. This is a common issue in many languages. Usually, you would store the value first and then remove like so:
local e
for i = 1, #heads do
if (heads[i][2] == id) then
e = i
end
end
if e then table.remove(heads, e) end
However, this solution is slow. Simply use the ID as key of your table:
local heads = {}
heads[1] = 50 -- push
heads[2] = 50
heads[3] = 50
heads[2] = nil -- remove
No need for unnecessary function calls and iterations.
According to 5.1 manual table.remove "Removes from table the element at position, shifting down other elements to close the space, if necessary"
size of heads (#heads) is calculated once before loop execution, when i==2 you call table.remove, and so size of the table shrinks to 2, and on next iteration you try index heads[3][2], but heads[3] is nil, therefore "attempt to index a nil value" error message.
As Andrew mentioned, for i = 1, #heads do will go to the original length of the list; if you shorten heads during the loop, then the final iteration(s) will read heads[i] and find only nil.
A simple way to fix this is to move backwards through the list, since removing an element only affects indices after the index you have removed from:
for i = #heads, 1, -1 do
if heads[i][2] == id then
table.remove(heads, i)
end
end
Note that in any case, this is O(n*d) complexity and could be very slow if you are deleting many elements from the list. And, as others pointed out, there's a O(1) approach where you use a map from v[1] => v instead.
To avoid problems caused by removing fields while iterating over an array, I've used a while loop with an index variable, which is incremented at the end of each iteration, but decremented when an index is removed. So, for instance, to remove all elements with an even index:
local t = { 1, 2, 3, 4, 5 }
local i = 1
while t[i] do
if t[i] % 2 == 0 then
table.remove(t, i)
i = i - 1
end
i = i + 1
end
This method allows you to iterate over array indices in ascending order.
Addressing this iteration, in case of many items that are to be removed, I'd rather go for
local result = {}
for _, v in ipairs(myTab) do
if v == 'nice' then
table.insert(result, v)
end
end
myTab = result
as table.remove is slow when shifting many elements.
How do you get the last index of a character in a string in Lua?
"/some/path/to/some/file.txt"
How do I get the index of the last / in the above string?
index = string.find(your_string, "/[^/]*$")
(Basically, find the position where the pattern "a forward slash, then zero or more things that aren't a forward slash, then the end of the string" occurs.)
This method is a bit more faster (it searches from the end of the string):
index = your_string:match'^.*()/'
Loops?!? Why would you need a loop for that? There's a 'reverse' native string function mind you, just apply it then get the first instance :) Follows a sample, to get the extension from a complete path:
function fileExtension(path)
local lastdotpos = (path:reverse()):find("%.")
return (path:sub(1 - lastdotpos))
end
You can do it in a single line of course, I split it in two for legibility's sake.
Here is a complete solution.
local function basename(path)
return path:sub(path:find("/[^/]*$") + 1)
end
local s = "/aa/bb/cc/dd/ee.txt"
local sep = "/"
local lastIndex = nil
local p = string.find(s, sep, 1)
lastIndex = p
while p do
p = string.find(s, sep, p + 1)
if p then
lastIndex = p
end
end
print(lastIndex)
You could continue find next value, until find nil,record last position
What I'm currently trying to do is make a table of email addresses (as keys) that hold person_records (as values). Where the person_record holds 6 or so things in it. The problem I'm getting is that when I try to assign the email address as a key to a table it complains and says table index is nil... This is what I have so far:
random_record = split(line, ",")
person_record = {first_name = random_record[1], last_name = random_record[2], email_address = random_record[3], street_address = random_record[4], city = random_record[5], state = random_record[6]}
email_table[person_record.email_address] = person_record
I wrote my own split function that basically takes a line of input and pulls out the 6 comma seperated values and stores them in a table (random_record)
I get an error when I try to say email_table[person_record.email_address] = person_record.
But when I print out person_record.email_address it's NOT nil, it prints out the string I stored in it.. I'm so confused.
function split(str, pat)
local t = {} -- NOTE: use {n = 0} in Lua-5.0
local fpat = "(.-)" .. pat
local last_end = 1
local s, e, cap = str:find(fpat, 1)
while s do
if s ~= 1 or cap ~= "" then
table.insert(t,cap)
end
last_end = e+1
s, e, cap = str:find(fpat, last_end)
end
if last_end <= #str then
cap = str:sub(last_end)
table.insert(t, cap)
end
return t
end
The following code is copy and pasted from your example and runs just fine:
email_table = {}
random_record = {"first", "second", "third"}
person_record = {first_name = random_record[1], last_name = random_record[1], email_address = random_record[1]}
email_table[person_record.email_address] = person_record
So your problem is in your split function.
BTW, Lua doesn't have "hashtables". It simply has "tables" which store key/value pairs. Whether these happen to use hashes or not is an implementation detail.
It looks like you iterating over some lines that have comma-separated data.
Looking at your split function, it stops as soon as there's no more separator (,) symbols in particular line to find. So feeding it anything with less than 3 ,-separated fields (for very common example: an empty line at end of file) will produce a table that doesn't go up to [3]. Addressing any empty table value will return you a nil, so person_record.email_address will be set to nil as well on the 2nd line of your code. Then, when you attempt to use this nil stored in person_record.email_address as an index to email_table in 3rd line, you will get the exact error you've mentioned.
I have the following function in Lua:
function iffunc(k,str,str1)
if k ~= 0 then
return str .. k .. (str1 or "")
end
end
This function allows me to check if value k is populated or not. I'm actually using it to determine if I want to display something that has zero value. My problem is this: I'm trying to concatenate a string of iffunc(), but since some of the values are 0, it returns an error of trying to concatenate a nil value. For instance:
levellbon = iffunc(levellrep["BonusStr"],"#wStr#r{#x111","#r}") .. iffunc(levellrep["BonusInt"],"#wInt#r{#x111","#r}") .. iffunc(levellrep["BonusWis"],"#wWis#r{#x111","#r}")
If any of the table values are 0, it'll return the error. I could easily put 'return 0' in the iffunc itself; however, I don't want a string of 000, either. So how can I work it where no matter which values are nil, I won't get that error? Ultimately, I'm going to do an iffunc on the levellbon variable to see if that's populated or not, but I've got that part figured out. I just need to get past this little hurdle right now. Thanks!
I'd do this:
function iffunc(k,str,str1)
if k == 0 then return "" end
return str .. k .. (str1 or "")
end
You should add an else statement in the function, where you return an empty string ("").