Reputation: 168239
Given a (multiline) string, where each line is separated by "\n"
and may not be necessarily of the same length, what is the best way to transpose it into another string as follows? Lines shorter than the longest one should be padded with space (right padding in terms of the original, or bottom padding in terms of the output). Applying the operation on a string twice should be idempotent modulo padding.
Input string
abc
def ghi
jk lm no
Output string
adj
bek
cf
l
gm
h
in
o
Upvotes: 1
Views: 3368
Reputation: 110745
Here are five approaches. (Yes, I got a bit carried away, but I find that trying to think of different ways to accomplish the same task is good exercise for the grey cells.)
#1
An uninteresting, brute-force method:
a = str.split("\n")
l = a.max_by(&:size).size
puts a.map { |b| b.ljust(l).chars }
.transpose
.map { |c| c.join.rstrip }.join("\n")
adj
bek
cf
l
gm
h
in
o
#2
This method and all that follow avoid the use of ljust
and transpose
, and make use of the fact that if e
is an empty array, e.shift
returns nil
and leaves e
an empty array. (Aside: I am often reaching for the non-existent method String#shift
. Here it would have avoided the need to convert each line to an array of characters.)
a = str.split("\n").map(&:chars)
a.max_by(&:size).size.times.map { a.map { |e| e.shift || ' ' }.join.rstrip }
#3
This and the remaining methods avoid the need to compute the length of the longest string:
a = str.split("\n").map(&:chars)
a_empty = Array(a.size, [])
[].tap { |b| b << a.map { |e| e.shift || ' ' }.join.rstrip while a != a_empty }
#4
This method makes use of Enumerator#lazy, which has been available since v2.0.
a = str.split("\n").map(&:chars)
(0..Float::INFINITY).lazy.map do |i|
a.each { |e| e.shift } if i > 0
a.map { |e| e.first || ' ' }.join.rstrip
end.take_while { c = a.any? { |e| !e.empty? } }.to_a
(I initially had a problem getting this to work, as I was not getting the element of the output (" o"
). The fix was adding the third line and changing the line that follows from a.map { |e| e.shift || ' ' }.join.rstrip
to what I have now. I mention this because it seems like it may be common problem when using lazy
.)
#5
Lastly, use recursion:
def recurse(a, b=[])
return b[0..-2] if a.last.empty?
b << a.map { |e| e.shift || ' ' }.join.rstrip
recurse(a, b)
end
a = str.split("\n").map(&:chars)
recurse(a)
Upvotes: 2
Reputation: 752
I would write it like this:
def transpose_text(text)
# split the text into lines
text = text.split("\n")
# find the length of the longest line
max_line_length = text.map(&:size).max
# pad each line with white space and convert them to character arrays
text.map! { |line| line.ljust(max_line_length).chars }
#transpose the character arrays and then join them all into one string
text.transpose.map(&:join).join("\n")
end
Upvotes: 0
Reputation: 42149
A play on what Chron did for an earlier version of ruby (e.g., 1.8.x). Example based on your original input that showed newline characters
str="abc\\n
def ghi\\n
jk lm no\\n"
def transpose s
lines = s.gsub("\\n","").split("\n")
longest = lines.map { |line| line.length }.max
(0..longest).map do |char_index|
lines.map { |line| line.split('')[char_index] || ' ' }.join
end * "\\n\n"
end
puts transpose(str)
Upvotes: 0
Reputation: 1127
I would write it like this:
def transpose s
lines = s.split(?\n)
longest = lines.map { |l| l.length }.max
(0..longest).map do |index|
lines.map { |l| l[index] || ' ' }.join
end * ?\n
end
Upvotes: 1
Reputation: 12782
This one works
s = "abc\ndef ghi\njk lm no\n"
s = s.split("\n")
s2 = ''
i = 0
while true
line = ''
s.each do |row|
line += (row[i] or ' ')
end
if line.strip == ''
break
end
s2 += line + "\n"
i += 1
end
puts s2
This one also works
s = "abc\ndef ghi\njk lm no\n"
s = s.split("\n")
maxlen = s.inject(0) {|m,r| m=[m, r.length].max}
s.map! {|r| r.ljust(maxlen).split(//)}
s = s.transpose.map {|r| r.join('')}.join("\n")
puts s
Upvotes: 0