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Join the elements of a list, separated by commas
Given the list
[Apple, Banana, Carrot] produce "Apple, Banana, Carrot"
perl
print join ', ', qw(Apple Banana Carrot);
# Longer and less efficient than join(), but illustrates
# Perl's foreach operator, which can be useful for
# less trivial problems with lists
@list = ('Apple', 'Banana', 'Carrot');
foreach $fruit (@list) {
print "$fruit,";
}
print "\n";
# Perl's foreach operator, which can be useful for
# less trivial problems with lists
@list = ('Apple', 'Banana', 'Carrot');
foreach $fruit (@list) {
print "$fruit,";
}
print "\n";
my @a = qw/Apple Banana Carrot/;
{
local $, = ", ";
print @a
}
print "\n";
{
local $, = ", ";
print @a
}
print "\n";
my @a = qw/Apple Banana Carrot/;
{
local $" = ", ";
print "@a\n";
}
{
local $" = ", ";
print "@a\n";
}
cpp
String^ result = String::Join(L", ", fruit->ToArray());
string fruits[] = {"Apple", "Banana", "Carrot"};
string result = boost::algorithm::join(fruits, ", ");
string result = boost::algorithm::join(fruits, ", ");
csharp
using System.Collections.Generic;
public class JoinEach {
public static void Main() {
var list = new List<string>() {"Apple", "Banana", "Carrot"};
System.Console.WriteLine( string.Join(", ", list.ToArray()) );
}
}
public class JoinEach {
public static void Main() {
var list = new List<string>() {"Apple", "Banana", "Carrot"};
System.Console.WriteLine( string.Join(", ", list.ToArray()) );
}
}
Join the elements of a list, in correct english
Create a function join that takes a List and produces a string containing an english language concatenation of the list. It should work with the following examples:
join(
join(
join(
join(
join(
[Apple, Banana, Carrot]) = "Apple, Banana, and Carrot"
join(
[One, Two]) = "One and Two"
join(
[Lonely]) = "Lonely"
join(
[]) = ""
perl
sub myjoin {
$_ = join ', ', @_;
s/, ([^,]+)$/ and $1/;
return $_;
}
# Note: I don't think this meets the spec --Geoff
$_ = join ', ', @_;
s/, ([^,]+)$/ and $1/;
return $_;
}
# Note: I don't think this meets the spec --Geoff
sub myjoin {
if ($#_ < 2) {
return join ' and ', @_;
} else {
return join(', ', @_[0..$#_-1]) . ' and ' . $_[-1];
}
}
# Note: I don't think this meets the spec --Geoff
if ($#_ < 2) {
return join ' and ', @_;
} else {
return join(', ', @_[0..$#_-1]) . ' and ' . $_[-1];
}
}
# Note: I don't think this meets the spec --Geoff
# Previous "myjoin()" responses don't meet the spec of including
# the final comma before the "and" if the list has more than
# two elements...this is one way to meet that spec...it may
# not be the most efficient...
sub AnotherMyJoin {
my @list = @_;
if ($#list == -1) {return}
elsif ($#list == 0) {return $list[0]}
elsif ($#list == 1) {return $list[0].' and '.$list[1]}
else {
return join(", ", @list[0..$#list - 1]) . ', and '. $list[$#list];
}
}
# the final comma before the "and" if the list has more than
# two elements...this is one way to meet that spec...it may
# not be the most efficient...
sub AnotherMyJoin {
my @list = @_;
if ($#list == -1) {return}
elsif ($#list == 0) {return $list[0]}
elsif ($#list == 1) {return $list[0].' and '.$list[1]}
else {
return join(", ", @list[0..$#list - 1]) . ', and '. $list[$#list];
}
}
# This is the long way, but it's kind of fun
# It illustrates the use of Perl's reverse()
# operator to work our way through the list
# elements backwards...I wrote this one before
# getting smart and looking at some of the other
# algorithms from the other languages. Still,
# it is only 12 lines of code vs 9 for my other
# solution if you disregard the comments.
sub myjoin {
my @list = reverse(@_); # Reverse original order of elements
my $retval;
# Make our exit here if we were passed an empty list
if ($#list == -1) {return}
# Loop through reversed elements in end-to-start order
for (0..$#list) {
# Add the reversed form of each element plus a space char
$retval .= reverse($list[$_]).' ';
# Add 'and' to lists with two or more elements
# placing it in between final and 'next to final'
$retval .= "dna " if ($#list > 0 and $_== 0);
# Add ',' to each element as long as there are more
# than two elements and the current element isn't the
# final element
$retval .= "," if ($#list > 1 and $_ != $#list);
}
# Remove what will end up as an extraneous leading space
chop($retval);
# Done looping, now reverse things back into correct order and return
$retval = reverse($retval);
return($retval);
}
# It illustrates the use of Perl's reverse()
# operator to work our way through the list
# elements backwards...I wrote this one before
# getting smart and looking at some of the other
# algorithms from the other languages. Still,
# it is only 12 lines of code vs 9 for my other
# solution if you disregard the comments.
sub myjoin {
my @list = reverse(@_); # Reverse original order of elements
my $retval;
# Make our exit here if we were passed an empty list
if ($#list == -1) {return}
# Loop through reversed elements in end-to-start order
for (0..$#list) {
# Add the reversed form of each element plus a space char
$retval .= reverse($list[$_]).' ';
# Add 'and' to lists with two or more elements
# placing it in between final and 'next to final'
$retval .= "dna " if ($#list > 0 and $_== 0);
# Add ',' to each element as long as there are more
# than two elements and the current element isn't the
# final element
$retval .= "," if ($#list > 1 and $_ != $#list);
}
# Remove what will end up as an extraneous leading space
chop($retval);
# Done looping, now reverse things back into correct order and return
$retval = reverse($retval);
return($retval);
}
# Yes, this doesn't meet the spec, the spec is flawed
# the serial comma (Oxford comma) is not required in a list
sub english_join {
return join(', ', @_[0..$#_-1])
. ($#_ ? ' and ' : '' )
. $_[-1];
}
# the serial comma (Oxford comma) is not required in a list
sub english_join {
return join(', ', @_[0..$#_-1])
. ($#_ ? ' and ' : '' )
. $_[-1];
}
cpp
Console::WriteLine(join(fruit));
string join(const vector<string> &s, int b=0)
{
switch (s.size() - b)
{
case 0: return "";
case 1: return s[b];
case 2: return s[b] + (s.size() > 2 ? "," : "") + " and " + s[b+1];
default: return s[b] + ", " + join(s, b+1);
}
}
{
switch (s.size() - b)
{
case 0: return "";
case 1: return s[b];
case 2: return s[b] + (s.size() > 2 ? "," : "") + " and " + s[b+1];
default: return s[b] + ", " + join(s, b+1);
}
}
csharp
using System.Collections.Generic;
using System.Linq;
public class CSharpListToEnglishList {
public string JoinAsEnglishList (List<string> words) {
switch (words.Count) {
case 0: return "";
case 1: return words[0];
case 2: return string.Format("{0} and {1}", words.ToArray());
default:
return JoinAsEnglishList( new List<string>() {
string.Join(", ", words.Take(words.Count - 1).ToArray()) + ",",
words.Last()
});
}
}
// Driver...
public static void Main() {
var joiner = new CSharpListToEnglishList();
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Apple", "Banana", "Carrot", "Orange" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Apple", "Banana", "Carrot" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "One", "Two" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Lonely" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>()) );
}
}
using System.Linq;
public class CSharpListToEnglishList {
public string JoinAsEnglishList (List<string> words) {
switch (words.Count) {
case 0: return "";
case 1: return words[0];
case 2: return string.Format("{0} and {1}", words.ToArray());
default:
return JoinAsEnglishList( new List<string>() {
string.Join(", ", words.Take(words.Count - 1).ToArray()) + ",",
words.Last()
});
}
}
// Driver...
public static void Main() {
var joiner = new CSharpListToEnglishList();
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Apple", "Banana", "Carrot", "Orange" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Apple", "Banana", "Carrot" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "One", "Two" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>() { "Lonely" }) );
System.Console.WriteLine(
joiner.JoinAsEnglishList(new List<string>()) );
}
}
Produce the combinations from two lists
Given two lists, produce the list of tuples formed by taking the combinations from the individual lists. E.g. given the letters
["a", "b", "c"] and the numbers [4, 5], produce the list: [["a", 4], ["b", 4], ["c", 4], ["a", 5], ["b", 5], ["c", 5]]
perl
@letters = qw(a b c);
@numbers = (4, 5);
@list = map { $number=$_; map [$_, $number], @letters; } @numbers;
@numbers = (4, 5);
@list = map { $number=$_; map [$_, $number], @letters; } @numbers;
@letters = qw(a b c);
@numbers = (4, 5);
for $number (@numbers) {
for $letter (@letters) {
push @list, [$letter, $number];
}
}
@numbers = (4, 5);
for $number (@numbers) {
for $letter (@letters) {
push @list, [$letter, $number];
}
}
cpp
Specialized::StringCollection^ combinations = gcnew Specialized::StringCollection;
for each(int number in numbers)
for each(String^ letter in letters)
combinations->Add(makeCombo(letter, number));
for each(int number in numbers)
for each(String^ letter in letters)
combinations->Add(makeCombo(letter, number));
string letters[] = { "a", "b", "c" };
int numbers[] = { 4, 5 };
list<pair<string,int> > combo;
for (int n = 0; n < sizeof numbers / sizeof *numbers; n++)
for (int l = 0; l < sizeof letters / sizeof *letters; l++)
combo.push_back(make_pair(letters[l], numbers[n]));
cout << combo << endl;
int numbers[] = { 4, 5 };
list<pair<string,int> > combo;
for (int n = 0; n < sizeof numbers / sizeof *numbers; n++)
for (int l = 0; l < sizeof letters / sizeof *letters; l++)
combo.push_back(make_pair(letters[l], numbers[n]));
cout << combo << endl;
csharp
using System.Collections.Generic;
public class ListCombiner {
public static void Main() {
var letters = new List<char>() { 'a', 'b', 'c' };
var numbers = new List<int>() { 1, 2, 3 };
// result is a list that contaings lists of objects
var result = new List<List<object>>();
foreach (var l in letters) {
foreach (var n in numbers) {
result.Add(new List<object>() { l, n });
}
}
}
}
public class ListCombiner {
public static void Main() {
var letters = new List<char>() { 'a', 'b', 'c' };
var numbers = new List<int>() { 1, 2, 3 };
// result is a list that contaings lists of objects
var result = new List<List<object>>();
foreach (var l in letters) {
foreach (var n in numbers) {
result.Add(new List<object>() { l, n });
}
}
}
}
From a List Produce a List of Duplicate Entries
Taking a list:
Write the code to produce a list of duplicates in the list:
["andrew", "bob", "chris", "bob"]
Write the code to produce a list of duplicates in the list:
["bob"]
perl
my @input = ("andrew", "bob", "chris", "bob", "bob");
my %input_count;
my @output = grep { $input_count{$_}++; $input_count{$_} == 2 } @input;
my %input_count;
my @output = grep { $input_count{$_}++; $input_count{$_} == 2 } @input;
cpp
vector<string> lst = { "andrew", "bob", "chris", "bob" };
vector<string> lst_no_dups;
vector<string> tmp;
vector<string> dups;
sort(lst.begin(), lst.end());
unique_copy(lst.begin(), lst.end(), back_inserter(lst_no_dups));
set_difference(lst.begin(), lst.end(),
lst_no_dups.begin(), lst_no_dups.end(),
back_inserter(tmp));
unique_copy(tmp.begin(), tmp.end(), back_inserter(dups));
cout << dups << endl;
vector<string> lst_no_dups;
vector<string> tmp;
vector<string> dups;
sort(lst.begin(), lst.end());
unique_copy(lst.begin(), lst.end(), back_inserter(lst_no_dups));
set_difference(lst.begin(), lst.end(),
lst_no_dups.begin(), lst_no_dups.end(),
back_inserter(tmp));
unique_copy(tmp.begin(), tmp.end(), back_inserter(dups));
cout << dups << endl;
list<string> lst = { "andrew", "bob", "chris", "bob" };
map<string,int> num_identical;
list<string> dups;
for (auto &s: lst)
num_identical[s]++;
for (auto &n: num_identical)
if (n.second > 1)
dups.push_back(n.first);
cout << dups << endl;
map<string,int> num_identical;
list<string> dups;
for (auto &s: lst)
num_identical[s]++;
for (auto &n: num_identical)
if (n.second > 1)
dups.push_back(n.first);
cout << dups << endl;
csharp
List<String> values = new List<string> {"andrew", "bob", "chris", "bob"};
var duplicates = values
.GroupBy(i => i)
.Where(j => j.Count() > 1)
.Select(s => s.Key);
foreach (var duplicate in duplicates)
{
Console.WriteLine(duplicate);
}
var duplicates = values
.GroupBy(i => i)
.Where(j => j.Count() > 1)
.Select(s => s.Key);
foreach (var duplicate in duplicates)
{
Console.WriteLine(duplicate);
}
