I have two strings, a and b, in Ruby.
a="scar"
b="cars"
What is the easiest way in Ruby to find whether a and b contain the same characters?
UPDATE
I am building an Anagram game ,so scar is an anagram of cars.So i want a way to compare a and b and come to conclusion that its an anagram
So c="carcass" should not be a match
You could do like this:
a = 'scar'
b = 'cars'
a.chars.sort == b.chars.sort
# => true
a = 'cars'
b = 'carcass'
a.chars.sort == b.chars.sort
# => false
Just for testing arrays vs string vs delete comparation. Assuming we compare strings with equal length.
In the real anagram search you need to sort first word a once. And then compare it to bunch of b's.
a="scar"
b="cars"
require 'benchmark'
n = 1000000
Benchmark.bm do |x|
x.report('string') { a = a.chars.sort.join; n.times do ; a == b.chars.sort.join ; end }
x.report('arrays') { a = a.chars.sort; n.times do ; a == b.chars.sort ; end }
end
The result:
user system total real
string 6.030000 0.010000 6.040000 ( 6.061088)
arrays 6.420000 0.010000 6.430000 ( 6.473158)
But, if you sort a each time (for delete we don't need to sort any word):
x.report('string') { n.times do ; a.chars.sort.join == b.chars.sort.join ; end }
x.report('arrays') { n.times do ; a.chars.sort == b.chars.sort ; end }
x.report('delete') { n.times do ; a.delete(b).empty? ; end }
The result is:
user system total real
string 11.800000 0.020000 11.820000 ( 11.989071)
arrays 11.210000 0.020000 11.230000 ( 11.263627)
delete 1.680000 0.000000 1.680000 ( 1.673979)
What is the easiest way in Ruby to find whether a and b contain the same characters?
As per the definitions of Anagram the below written code should work :
a="scar"
b="cars"
a.size == b.size && a.delete(b).empty?
require 'set'
Set.new(a.chars) == Set.new(b.chars)
updated to take into account comment from sawa
Related
Looking for direction on how to chunk and compare two large text files using ruby. Any help is appreciated. Something like 100 lines at a time.
tried:
file(file1).foreach.each_slice(100) do |lines|
pp lines
end
getting confused how to include the second file to this loop.
CHUNK_SIZE = 256 # bytes
def same? path1, path2
return false unless [path1, path2].map { |f| File.size f }.reduce &:==
f1, f2 = [path1, path2].map { |f| File.new f }
loop do
s1, s2 = [f1, f2].map { |f| f.read(CHUNK_SIZE) }
break false if s1 != s2
break true if s1.nil? || s1.length < CHUNK_SIZE
end
ensure
[f1, f2].each &:close
end
UPD: credits for fixed typo and file size comparison goes to #tadman.
Just "Process two files at the same time in Ruby" and compare by chunks, like this:
f1 = File.open('file1.txt', 'r')
f2 = File.open('file2.txt', 'r')
f1.each_slice(10).zip(f2.each_slice(10)).each do |line1, line2|
return false unless line1 == line2
end
return true
Or, as suggested by #meagar (in this case line by line):
f1.each_line.zip(f2.each_line).all? { |a,b| a == b }
This will return true if files identical.
Just compare those files line by line:
def same_file?(path1, path2)
file1 = File.open(path1)
file2 = File.open(path2)
return true if File.absolute_path(path1) == File.absolute_path(path2)
return false unless file1.size == file2.size
enum1 = file1.each
enum2 = file2.each
loop do
# It's a mystery that the loop really ends
# when any of the 2 files has nothing to read
return false unless enum1.next == enum2.next
end
return true
ensure
file1.close
file2.close
end
I did my homework and found in the Kernel#loop documentation:
StopIteration raised in the block breaks the loop. In this case, loop returns the "result" value stored in the exception.
And, in the Enumerator#next documentation:
When the position reached at the end, StopIteration is raised.
So the mystery is no longer a mystery for me.
Here's another one, the approach is similar to mudasobwa's answer:
def same?(file_1, file_2)
return true if File.identical?(file_1, file_2)
return false unless File.size(file_1) == File.size(file_2)
buf_size = 2 ** 15 # 32 K
buf_1 = ''
buf_2 = ''
File.open(file_1) do |f1|
File.open(file_2) do |f2|
while f1.read(buf_size, buf_1) && f2.read(buf_size, buf_2)
return false unless buf_1 == buf_2
end
end
end
true
end
In the first two lines perform quick checks for identical files (e.g. hard and soft links) and same size using File.identical? and File.size.
File.open opens each file in read-only mode. The while loop then keeps calling read to read 32K chunks from each file into the buffers buf_1 and buf_2 until EOF. If the buffers differ, false is returned. Otherwise, i.e. without encountering any differences, true is returned.
To determine if two files have the exact same content, without comparing the actual content of the same chunk of each file, you can use a checksum function that turns the data into a hash string in a deterministic way. And while you have to read the contents to checksum it, you can get checksums for each slice, and end up with an array of checksums for each file.
You can then compare the collection of checksums. If the two files have the exact same content, the two collections will be equal.
require 'digest/md5'
hashes1 = File.foreach('./path_to_file').each_slice(100).map do |slice|
Digest::MD5.hexdigest(slice)
end
hashes2 = File.read('./path_to_duplicate').each_slice(100).map do |slice|
Digest::MD5.hexdigest(slice)
end
hashes1.join == hashes2.join
#=> true, meaning the two files contain the same content
Benchmark time
(Matt's answer is not included because I couldn't get it working)
Results 1 KB file size (N = 10000)
user system total real
aetherus 0.510000 0.300000 0.810000 ( 0.823201)
meagar 0.350000 0.160000 0.510000 ( 0.512755)
mudasobwa 0.290000 0.200000 0.490000 ( 0.500831)
stefan 0.150000 0.160000 0.310000 ( 0.312743)
yevgeniy_anfilofyev 0.320000 0.170000 0.490000 ( 0.497157)
Results 1 MB file size (N = 100)
user system total real
aetherus 1.540000 0.110000 1.650000 ( 1.667937)
meagar 1.170000 0.130000 1.300000 ( 1.310278)
mudasobwa 1.470000 0.830000 2.300000 ( 2.313481)
stefan 0.010000 0.030000 0.040000 ( 0.045577)
yevgeniy_anfilofyev 0.570000 0.100000 0.670000 ( 0.677226)
Results 1 GB file size (N = 1)
user system total real
aetherus 15.570000 0.920000 16.490000 ( 16.525826)
meagar 24.170000 1.910000 26.080000 ( 26.190057)
mudasobwa 16.260000 8.160000 24.420000 ( 24.471977)
stefan 0.120000 0.330000 0.450000 ( 0.443074)
yevgeniy_anfilofyev 12.940000 1.310000 14.250000 ( 14.295736)
Notes
mudasobwa's code runs significantly faster with larger CHUNK_SIZE
with identical chunk sizes, stefan's code seems to be ~2x faster than mudasobwa's code
"fastest" chunk size is somewhere between 16 K and 512 K
I couldn't use fruity because the 1 GB test would have taken too long
Code
def aetherus_same?(f1, f2)
enum1 = f1.each
enum2 = f2.each
loop do
return false unless enum1.next == enum2.next
end
return true
end
def meagar_same?(f1, f2)
f1.each_line.zip(f2.each_line).all? { |a,b| a == b }
end
CHUNK_SIZE = 256 # bytes
def mudasobwa_same?(f1, f2)
loop do
s1, s2 = [f1, f2].map { |f| f.read(CHUNK_SIZE) }
break false if s1 != s2
break true if s1.nil? || s1.length < CHUNK_SIZE
end
end
def stefan_same?(f1, f2)
buf_size = 2 ** 15 # 32 K
buf_1 = ''
buf_2 = ''
while f1.read(buf_size, buf_1) && f2.read(buf_size, buf_2)
return false unless buf_1 == buf_2
end
true
end
def yevgeniy_anfilofyev_same?(f1, f2)
f1.each_slice(10).zip(f2.each_slice(10)).each do |line1, line2|
return false unless line1 == line2
end
return true
end
FILE1 = ARGV[0]
FILE2 = ARGV[1]
N = ARGV[2].to_i
def with_files
File.open(FILE1) { |f1| File.open(FILE2) { |f2| yield f1, f2 } }
end
require 'benchmark'
Benchmark.bm(19) do |x|
x.report('aetherus') { N.times { with_files { |f1, f2| aetherus_same?(f1, f2) } } }
x.report('meagar') { N.times { with_files { |f1, f2| meagar_same?(f1, f2) } } }
x.report('mudasobwa') { N.times { with_files { |f1, f2| mudasobwa_same?(f1, f2) } } }
x.report('stefan') { N.times { with_files { |f1, f2| stefan_same?(f1, f2) } } }
x.report('yevgeniy_anfilofyev') { N.times { with_files { |f1, f2| yevgeniy_anfilofyev_same?(f1, f2) } } }
end
I want to create a regexp to specify words between parenthesis. For example, I have a string like this:
"something(a,b)"
and
"something(c, d)"
and I want to extract the letters from between the parentheses.
In the first string I want to get an array ['a','b']. In the second, I want the array ['c','d'].
I have following method:
def suffixes(t)
(t.scan /\((\w+),(\w+)\)/).flatten
end
but this works only for the first case. In the second variant I have:
def suffixes(t)
(t.scan /\((\w+),[\s](\w+)\)/).flatten
end
But this works only for the second case. I don't know what regexp will operate in both cases.
You can use:
def suffixes(t)
(t.scan /\((\w+)\s*,\s*(\w+)\)/).flatten
end
\s* will match 0 or more spaces before and after comma.
Make the inbetween \s as optional.
def suffixes(t)
(t.scan /\((\w+),\s?(\w+)\)/).flatten
end
? after the \s would turn the space to optional (0 or 1).
I would suggest you to distinguish "scanning" for the text between parentheses and "splitting" the result by comma:
s = "something(c, d)"
s.match( /\((.+)\)/ )[1] # found text between parentheses
.split(/,/) # split the result by comma
.map(&:strip) # stripped the values
It’s more Ruby-like, in my understanding. Hope it helps.
UPD Thanks #theTinMan, there are two possibilities to improve an answer. First of all, s[/\((.+)\)/, 1] looks better ans executes faster than s.match( /\((.+)\)/ )[1]. Secondary, splitting by string is faster than splitting by regexp. The summing up:
s = "something(c, d)"
s[ /\((.+)\)/, 1 ] # found text between parentheses
.split(',') # split the result by comma
.map(&:strip) # stripped the values
Proof:
require 'benchmark'
n = 1_000_000
s = "something(c, d)"
Benchmark.bm do |x|
x.report { n.times { s.match( /\((.+)\)/ )[1].split(/,/).map(&:strip) } }
x.report { n.times { s.match( /\((.+)\)/ )[1].split(',').map(&:strip) } }
x.report { n.times { s[/\((.+)\)/, 1].split(/,/).map(&:strip) } }
x.report { n.times { s[/\((.+)\)/, 1].split(',').map(&:strip) } }
end
# user system total real
# 3.590000 0.000000 3.590000 ( 3.598151)
# 3.030000 0.000000 3.030000 ( 3.028137)
# 2.940000 0.000000 2.940000 ( 2.942490)
# 2.180000 0.000000 2.180000 ( 2.182447)
\((\w+)|(?!^)\G\s*,\s*(\w+)
Try this.This will work for all arguments.See demo.
https://regex101.com/r/vN3sH3/27
I'm scrubbing large data files (+1MM comma-separated rows). An example row might look like this:
#row = "123456789,11122,CustomerName,2014-01-31,2014-02-01,RemoveThisEntry,R,SKUInfo,05-MAR-14 05:50:24,SourceID,RemoveThisEntryToo,TransactionalID"
Certain columns must be removed from it, after which the row should look like this:
#row = "123456789,11122,CustomerName,2014-01-31,2014-02-01,R,SKUInfo,05-MAR-14 05:50:24,SourceID,TransactionalID"
QUESTION 1: If I convert a row of data into an Array, which method is preferred for removing elements: Array#delete_at or Array#slice!? I'd like to know which is the more idiomatic option. Performance is a consideration here, and I'm on a Windows machine.
def remove_bad_columns
ary = #row.split(",")
ary.delete_at(10)
ary.delete_at(5)
#row = ary.join(",")
end
QUESTION 2: I was wondering if one of these methods was implemented using the other. How can I see how the methods are built in ruby? (How for is implemented using each, for example.)
I suggest you use Array#values_at rather than delete_at or slice!:
def remove_vals(str, *indices)
ary = str.split(",")
v = (0...ary.size).to_a - indices
ary.values_at(*v).join(",")
end
#row = "123456789,11122,CustomerName,2014-01-31,2014-02-01,RemoveThisEntry," +
"R,SKUInfo,05-MAR-14 05:50:24,SourceID,RemoveThisEntryToo,TransactionalID"
#row = remove_vals(#row, 5, 10)
#=> "123456789,11122,CustomerName,2014-01-31,2014-02-01,R,SKUInfo," +
# "05-MAR-14 05:50:24,SourceID,TransactionalID"
Array#values_at has the advantage over the other two methods that you don't have to worry about the order in which the elements are removed.
The efficiency of this method is not significantly different than the other two. If #spickermann would like to add it to his benchmarks, he could use this:
def values_at
ary = array.split(",")
v = (0...ary.size).to_a - [5,10]
#row = ary.values_at(*v).join(",")
end
There is not really a difference in performance. I would prefer delete_at because that reads nicer.
require 'benchmark'
def array
"123456789,11122,CustomerName,2014-01-31,2014-02-01,RemoveThisEntry,R,SKUInfo,05-MAR-14 05:50:24,SourceID,RemoveThisEntryToo,TransactionalID"
end
def delete_at
ary = array.dup.split(",")
ary.delete_at(10)
ary.delete_at(5)
#row = ary.join(",")
end
def slice!
ary = array.dup.split(",")
ary.slice!(10)
ary.slice!(5)
#row = ary.join(",")
end
require 'benchmark'
n = 1_000_000
Benchmark.bmbm(15) do |x|
x.report("delete_at :") { n.times do; delete_at; end }
x.report("slice! :") { n.times do; slice! ; end }
end
# Rehearsal ---------------------------------------------------
# delete_at : 4.560000 0.000000 4.560000 ( 4.566496)
# slice! : 4.580000 0.010000 4.590000 ( 4.576767)
# ------------------------------------------ total: 9.150000sec
#
# user system total real
# delete_at : 4.500000 0.000000 4.500000 ( 4.505638)
# slice! : 4.600000 0.000000 4.600000 ( 4.613447)
What is the most efficient way to check if two hashes h1 and h2 have the same set of keys disregarding the order? Could it be made faster or more concise with close efficiency than the answer that I post?
Alright, let's break all rules of savoir vivre and portability. MRI's C API comes into play.
/* Name this file superhash.c. An appropriate Makefile is attached below. */
#include <ruby/ruby.h>
static int key_is_in_other(VALUE key, VALUE val, VALUE data) {
struct st_table *other = ((struct st_table**) data)[0];
if (st_lookup(other, key, 0)) {
return ST_CONTINUE;
} else {
int *failed = ((int**) data)[1];
*failed = 1;
return ST_STOP;
}
}
static VALUE hash_size(VALUE hash) {
if (!RHASH(hash)->ntbl)
return INT2FIX(0);
return INT2FIX(RHASH(hash)->ntbl->num_entries);
}
static VALUE same_keys(VALUE self, VALUE other) {
if (CLASS_OF(other) != rb_cHash)
rb_raise(rb_eArgError, "argument needs to be a hash");
if (hash_size(self) != hash_size(other))
return Qfalse;
if (!RHASH(other)->ntbl && !RHASH(other)->ntbl)
return Qtrue;
int failed = 0;
void *data[2] = { RHASH(other)->ntbl, &failed };
rb_hash_foreach(self, key_is_in_other, (VALUE) data);
return failed ? Qfalse : Qtrue;
}
void Init_superhash(void) {
rb_define_method(rb_cHash, "same_keys?", same_keys, 1);
}
Here's a Makefile.
CFLAGS=-std=c99 -O2 -Wall -fPIC $(shell pkg-config ruby-1.9 --cflags)
LDFLAGS=-Wl,-O1,--as-needed $(shell pkg-config ruby-1.9 --libs)
superhash.so: superhash.o
$(LINK.c) -shared $^ -o $#
An artificial, synthetic and simplistic benchmark shows what follows.
require 'superhash'
require 'benchmark'
n = 100_000
h1 = h2 = {a:5, b:8, c:1, d:9}
Benchmark.bm do |b|
# freemasonjson's state of the art.
b.report { n.times { h1.size == h2.size and h1.keys.all? { |key| !!h2[key] }}}
# This solution
b.report { n.times { h1.same_keys? h2} }
end
# user system total real
# 0.310000 0.000000 0.310000 ( 0.312249)
# 0.050000 0.000000 0.050000 ( 0.051807)
Combining freemasonjson's and sawa's ideas:
h1.size == h2.size and (h1.keys - h2.keys).empty?
Try:
# Check that both hash have the same number of entries first before anything
if h1.size == h2.size
# breaks from iteration and returns 'false' as soon as there is a mismatched key
# otherwise returns true
h1.keys.all?{ |key| !!h2[key] }
end
Enumerable#all?
worse case scenario, you'd only be iterating through the keys once.
Just for the sake of having at least a benchmark on this question...
require 'securerandom'
require 'benchmark'
a = {}
b = {}
# Use uuid to get a unique random key
(0..1_000).each do |i|
key = SecureRandom.uuid
a[key] = i
b[key] = i
end
Benchmark.bmbm do |x|
x.report("#-") do
1_000.times do
(a.keys - b.keys).empty? and (a.keys - b.keys).empty?
end
end
x.report("#&") do
1_000.times do
computed = a.keys & b.keys
computed.size == a.size
end
end
x.report("#all?") do
1_000.times do
a.keys.all?{ |key| !!b[key] }
end
end
x.report("#sort") do
1_000.times do
a_sorted = a.keys.sort
b_sorted = b.keys.sort
a == b
end
end
end
Results are:
Rehearsal -----------------------------------------
#- 1.000000 0.000000 1.000000 ( 1.001348)
#& 0.560000 0.000000 0.560000 ( 0.563523)
#all? 0.240000 0.000000 0.240000 ( 0.239058)
#sort 0.850000 0.010000 0.860000 ( 0.854839)
-------------------------------- total: 2.660000sec
user system total real
#- 0.980000 0.000000 0.980000 ( 0.976698)
#& 0.560000 0.000000 0.560000 ( 0.559592)
#all? 0.250000 0.000000 0.250000 ( 0.251128)
#sort 0.860000 0.000000 0.860000 ( 0.862857)
I have to agree with #akuhn that this would be a better benchmark if we had more information on the dataset you are using. But that being said, I believe this question really needed some hard fact.
It depends on your data.
There is no general case really. For example, generally retrieving the entire keyset at once is faster than checking inclusion of each key seperately. However, if in your dataset, the keysets differ more often than not, then a slower solution which fails faster might be faster. For example:
h1.size == h2.size and h1.keys.all?{|k|h2.include?(k)}
Another factor to consider is the size of your hashes. If they are big a solution with higher setup cost, like calling Set.new, might pay off, if however they are small, it won't:
h1.size == h2.size and Set.new(h1.keys) == Set.new(h2.keys)
And if you happen to compare the same immutable hashes over and over again, it would definitely pay off to cache the results.
Eventually only a benchmark will tell, but, to write a benchmark, we'd need to know more about your use case. For sure, testing a solution with synthetic data (as for example, randomly generated keys) will not be representative.
This is my try:
(h1.keys - h2.keys).empty? and (h2.keys - h1.keys).empty?
Here is my solution:
class Hash
# doesn't check recursively
def same_keys?(compare)
if compare.class == Hash
if self.size == compare.size
self.keys.all? {|s| compare.key?(s)}
else
return false
end
else
nil
end
end
end
a = c = { a: nil, b: "whatever1", c: 1.14, d: true }
b = { a: "foo", b: "whatever2", c: 2.14, "d": false }
d = { a: "bar", b: "whatever3", c: 3.14, }
puts a.same_keys?(b) # => true
puts a.same_keys?(c) # => true
puts a.same_keys?(d) # => false
puts a.same_keys?(false).inspect # => nil
puts a.same_keys?("jack").inspect # => nil
puts a.same_keys?({}).inspect # => false
I believe I have a good answer to this issue, but I wanted to make sure ruby-philes didn't have a much better way to do this.
Basically, given an input string, I would like to convert the string to an integer, where appropriate, or a float, where appropriate. Otherwise, just return the string.
I'll post my answer below, but I'd like to know if there is a better way out there.
Ex:
to_f_or_i_or_s("0523.49") #=> 523.49
to_f_or_i_or_s("0000029") #=> 29
to_f_or_i_or_s("kittens") #=> "kittens"
I would avoid using regex whenever possible in Ruby. It's notoriously slow.
def to_f_or_i_or_s(v)
((float = Float(v)) && (float % 1.0 == 0) ? float.to_i : float) rescue v
end
# Proof of Ruby's slow regex
def regex_float_detection(input)
input.match('\.')
end
def math_float_detection(input)
input % 1.0 == 0
end
n = 100_000
Benchmark.bm(30) do |x|
x.report("Regex") { n.times { regex_float_detection("1.1") } }
x.report("Math") { n.times { math_float_detection(1.1) } }
end
# user system total real
# Regex 0.180000 0.000000 0.180000 ( 0.181268)
# Math 0.050000 0.000000 0.050000 ( 0.048692)
A more comprehensive benchmark:
def wattsinabox(input)
input.match('\.').nil? ? Integer(input) : Float(input) rescue input.to_s
end
def jaredonline(input)
((float = Float(input)) && (float % 1.0 == 0) ? float.to_i : float) rescue input
end
def muistooshort(input)
case(input)
when /\A\s*[+-]?\d+\.\d+\z/
input.to_f
when /\A\s*[+-]?\d+(\.\d+)?[eE]\d+\z/
input.to_f
when /\A\s*[+-]?\d+\z/
input.to_i
else
input
end
end
n = 1_000_000
Benchmark.bm(30) do |x|
x.report("wattsinabox") { n.times { wattsinabox("1.1") } }
x.report("jaredonline") { n.times { jaredonline("1.1") } }
x.report("muistooshort") { n.times { muistooshort("1.1") } }
end
# user system total real
# wattsinabox 3.600000 0.020000 3.620000 ( 3.647055)
# jaredonline 1.400000 0.000000 1.400000 ( 1.413660)
# muistooshort 2.790000 0.010000 2.800000 ( 2.803939)
def to_f_or_i_or_s(v)
v.match('\.').nil? ? Integer(v) : Float(v) rescue v.to_s
end
A pile of regexes might be a good idea if you want to handle numbers in scientific notation (which String#to_f does):
def to_f_or_i_or_s(v)
case(v)
when /\A\s*[+-]?\d+\.\d+\z/
v.to_f
when /\A\s*[+-]?\d+(\.\d+)?[eE]\d+\z/
v.to_f
when /\A\s*[+-]?\d+\z/
v.to_i
else
v
end
end
You could mash both to_f cases into one regex if you wanted.
This will, of course, fail when fed '3,14159' in a locale that uses a comma as a decimal separator.
Depends on security requirements.
def to_f_or_i_or_s s
eval(s) rescue s
end
I used this method
def to_f_or_i_or_s(value)
return value if value[/[a-zA-Z]/]
i = value.to_i
f = value.to_f
i == f ? i : f
end
CSV has converters which do this.
require "csv"
strings = ["0523.49", "29","kittens"]
strings.each{|s|p s.parse_csv(converters: :numeric).first}
#523.49
#29
#"kittens"
However for some reason it converts "00029" to a float.