hello, can you tell me how to make a better function that performs line feeds by spaces, than what I did with my natural science education, i, j, l and += 1 ?
function feed_lines(str::String, l::Int)
chars = collect(str)
i = findfirst(' ', str)
j = 1
for (index, char) in pairs(chars)
if char == ' '
i = index
end
if j > l
chars[i] = '\n'
j = index - i
end
j += 1
end
return join(chars)
end
although it seems not so bad
Could you explain what is the role of integer l?
In my limited understanding the following seems to produce similar results:
feed_lines2(str::String) = join(split(str),"\n")
this is line length, if line length > l then replace last space (which < l) to â\nâ
Could you provide a minimum working example (MWE) to illustrate what it is supposed to do?
sure
julia> function feed_lines(str::String, l::Int)
chars = collect(str)
i = findfirst(' ', str)
j = 1
for (index, char) in pairs(chars)
if char == ' '
i = index
end
if j > l
chars[i] = '\n'
j = index - i
end
j += 1
end
return join(chars)
end
feed_lines (generic function with 1 method)
julia> feed_lines("""hello, can you tell me how to make a better function that performs line feeds by spaces, than what I did with my natural science education, i, j, l and += 1 ?""", 30) |> println
hello, can you tell me how to
make a better function that
performs line feeds by spaces,
than what I did with my
natural science education, i,
j, l and += 1 ?
Just an idea (not quite doing what you ask for):
julia> join(join.(Iterators.partition(s, 30) |> collect, ""), "\n") |> println
hello, can you tell me how to
make a better function that pe
rforms line feeds by spaces, t
han what I did with my natural
science education, i, j, l an
d += 1 ?
where s is your example string.
(Do we have an efficient âword iteratorâ somewhere?)
This is rather compact:
julia> function feed_lines2(str, l)
line_breaks = findnext.(' ', str, l:l:length(str))
join([i in line_breaks ? '\n' : c for (i,c) in pairs(str)])
end
feed_lines2 (generic function with 1 method)
julia> feed_lines2("""hello, can you tell me how to make a better function that performs line feeds by spaces, than what I did with my natural science education, i, j, l and += 1 ?""", 30) |> println
hello, can you tell me how to
make a better function that performs
line feeds by spaces, than
what I did with my natural
science education, i, j, l and
+= 1 ?
However, I think it will be (a bit) less efficient than your variant.
A version that uses SubString and also returns the number of rows printed:
function feed_lines3(str, n)
ix = [1; findall(' ', str); length(str)]
j1 = zero(ix)
j2 = zero(ix)
nr = 1
j1[1] = 1
for i in 2:length(ix)
if ix[i] - j1[nr] > n
j2[nr] = ix[i-1]
nr += 1
j1[nr] = ix[i-1] + 1
elseif i == length(ix)
j2[nr] = ix[end]
end
end
join(SubString.(str, view(j1,1:nr), view(j2,1:nr)), "\n"), nr
end
println(feed_lines3(str, 30)[1])
# result:
hello, can you tell me how to
make a better function that
performs line feeds by spaces,
than what I did with my
natural science education, i,
j, l and += 1 ?
This version should be further optimized, as only one j array seems necessary.
FWIW, a simpler alternative and a bit faster:
function feed_lines4(str, n)
ix = [1; findall(' ', str); length(str)]
nx = length(ix)
j = 1;
s2 = ""
for i in 2:nx
if ix[i] - j > n
s2 = s2 * view(str,j:ix[i-1]) * "\n"
j = ix[i-1] + 1
elseif i == nx
s2 = s2 * view(str,j:ix[nx]) * "\n"
end
end
return s2
end
println(feed_lines4(str,30))
# result:
hello, can you tell me how to
make a better function that
performs line feeds by spaces,
than what I did with my
natural science education, i,
j, l and += 1 ?