I am having troubles how to start creating program which counts letters in words in .txt file, and print it if letter number is bigger than 20.. Does someone know how to start or have some similar program to help me..Thank you
In general,I start programs by:
#include <iostream>
void Pause();
int main()
{
std::cout << "My program\n";
Pause();
return 0;
}
void Pause()
{
std::cout << "\n\nPaused. Press ENTER to continue.\n";
std::cin.ignore(100000, '\n');
}
After I get the above working, I add content.
Assuming the ASCII encoding, you could make some assumptions:
int counts[256] = {0};
char c;
while (my_file.read(&c, 1))
{
counts[c]++;
}
The above loop will count all occurrences of the letters in the text file. The next task would be to print the data:
for (int i = ' ' + 1; i < 127; ++i)
{
std::cout << "'" << ((char) i) << "': " << counts[i] << \n;
}
This will print out the counts for all printable characters, starting with the character after a space.
This is a popular beginning assignment, so you should be able to find more examples on the internet.
Edit 1: For counts of words
The std::string and std::map are your friends.
std::string word;
std::map<std::string, int> database;
while (my_file >> word)
{
std::map<std::string, int>::iterator iter;
iter = database.find(word);
if (iter != database.end())
{
map[word]++;
}
else
{
map[word] = 1;
}
}
The above loop builds a database of [word, count] pairs. If the word exists, the count is incremented.
The next part is to iterate through the database and only print the words that have a length > 20. This is left as an exercise for the reader.
Related
I need to write a sentence in Pig Latin form and I am almost done it successfuly except for 1 case and I almost give up
for example :
If my word starts at a\e\o\u\i the word will look like easy -> easyway , apple -> appleway
and if it doesnt start with a letter that I wrote above
it will look like that: box -> oxbay , king -> ingkay
I succeed with the bolded part but in the first part with a\e\o\u\i letter at the beginning , I dont know where to put the w before and need some help with it
This is my code , thanks in advance
#include <iostream>
//Since those are used in ALL function of program, it wont hurt to set it to global
//Else it is considered EVIL to declare global variables
const int maxLine = 100;
char phraseLine[maxLine] = { '\0' };
void pigLatinString();
using namespace std;
void main()
{
// Displayed heading of program
cout << "* You will be prompted to enter a string of *" << endl;
cout << "* words. The string will be converted into *" << endl;
cout << "* Pig Latin and the results displayed. *" << endl;
cout << "* Enter as many strings as you would like. *" << endl;
//prompt the user for a group of words or press enter to quit
cout << "Please enter a word or group of words. (Press enter to quit)\n";
cin.getline(phraseLine, 100, '\n');
cout << endl;
// This is the main loop. Continue executing until the user hits 'enter' to quit.
while (phraseLine[0] != '\0')
{
// Display the word (s) entered by the user
cout << "You entered the following: " << phraseLine << endl;
// Display the word (s) in Pig Latin
cout << "The same phrase in Pig latin is: ";
pigLatinString();
cout << endl;
//prompt the user for a group of words or press enter to quit
cout << "Please enter a word or group of words. (Press enter to quit)\n";
cin.getline(phraseLine, 100, '\n');
}
return;
}
void pigLatinString() //phraseLine is a cstring for the word, maxline is max length of line
{ //variable declarations
char tempConsonant[10];
tempConsonant[0] = '\0';
int numberOfConsonants = 0;
char previousCharacter = ' ';
char currentCharacter = ' ';
bool isInWord = 0;
// for loop checking each index to the end of whatever is typed in
for (int i = 0; i < maxLine; i++)
{
//checking for the end of the phraseline
if (phraseLine[i] == '\0')
{//checking to see if it's in the word
if (isInWord)
{//checking to see that there wasn't a space ahead of the word and then sending the cstring + ay to the console
if (previousCharacter != ' ')
cout << tempConsonant << "ay" << endl;
}
return;
}
// this covers the end of the word condition
if (isInWord)
{// covers the condition of index [i] being the space at the end of the word
if (phraseLine[i] == ' ')
{
// spits out pig latin word, gets you out of the word, flushes the temp consonants array and resets the # of consonants to 0
cout << tempConsonant << "ay";
isInWord = 0;
tempConsonant[0] = '\0';
numberOfConsonants = 0;
}
cout << phraseLine[i] ;
}
else
{//this covers for the first vowel that makes the switch
if (phraseLine[i] != ' ')
{// sets the c string to what is in the phraseline at the time and makes it capitalized
char currentCharacter = phraseLine[i];
currentCharacter = toupper(currentCharacter);
// this takes care of the condition that currentCharacter is not a vowel
if ((currentCharacter != 'A') && (currentCharacter != 'E') &&
(currentCharacter != 'I') && (currentCharacter != 'O') && (currentCharacter != 'U'))
//this sets the array to temporarily hold the consonants for display before the 'ay'
{//this sets the null operator at the end of the c string and looks for the next consonant
tempConsonant[numberOfConsonants] = phraseLine[i];
tempConsonant[numberOfConsonants + 1] = '\0';
numberOfConsonants++;
}
else
{// this sets the boolean isInWord to true and displays the phraseline
isInWord = 1;
cout << phraseLine[i];
}
}
else
{
cout << phraseLine[i] ;
}
}
previousCharacter = phraseLine[i];
}
return;
}
You have two conditions to consider. if your word starts with a vowel, just add "way" to the end of the word, else move the first letter and add "ay" to the end.
This is a task that can be made a lot simpler by using std::string instead of C-strings. This is because you are now no longer concerned with exceeding your length or losing the null character. It also allows easier access to the Standard Library algorithms.
#include <algorithm>
#include <iostream>
#include <string>
std::string make_pig_latin(const std::string& word) {
std::string vowels("aeiou");
std::string newWord(word);
if (newWord.find_first_not_of(vowels) == 0) {
// Word starts with a consanant
std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
newWord += "ay";
} else {
newWord += "way";
}
return newWord;
}
int main() {
std::cout << make_pig_latin("apple") << '\n'
<< make_pig_latin("box") << '\n'
<< make_pig_latin("king") << '\n'
<< make_pig_latin("easy") << '\n';
}
The function above highlights how you can go about structuring your conversion. You just need to know if your word starts with a vowel or not, and take the appropriate action.
Output:
appleway
oxbay
ingkay
easyway
I did not get the impression that you have to care about words like 'phone'.
Looking through your code, you should try to do a better job at separating your concerns. Pig Latin is easier done one word at a time, but you have string splitting code and a lot of "not Pig Latin" code in your Pig Latin function. Your main can handle getting input. You should probably have a separate function to break the line up into individual words, using std::vector to hold the words would be best since it can grow on demand and doesn't have to know a specific capacity up front. You then iterate through your array of words and translate them individually. Depending on what your actual requirements are, it's possible that you don't even have to store the translated words, just print them directly to the screen.
Here's the same program, but now it can separate words. Note how the pig latin function doesn't have to change (much, I added upper-case vowels just because I didn't want to bothered converting words) in order for the added functionality to be added.
#include <algorithm>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>
std::string make_pig_latin(const std::string& word) {
std::string vowels("aeiouAEIOU");
std::string newWord(word);
if (newWord.find_first_not_of(vowels) == 0) {
// Word starts with a consanant
std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
newWord += "ay";
} else {
newWord += "way";
}
return newWord;
}
int main() {
std::string phrase(
"A sentence where I say words like apple box easy king and ignore "
"punctuation");
std::istringstream sin(phrase);
std::vector<std::string> words(std::istream_iterator<std::string>(sin), {});
for (auto i : words) {
std::cout << make_pig_latin(i) << ' ';
}
}
Output:
Away entencesay hereway Iway aysay ordsway ikelay appleway oxbay easyway ingkay andway ignoreway unctuationpay
I am currently self-studying C++ with Schaum's outline book (which covers mostly C contents, or so I've been told, but whatever) and I have encountered some trouble with problem 9.8.
You are supposed to count the number of appearances of every different word in a given c++ string, for which I assumed each word was separated from the next one by a white space, a newline or a dot or coma (followed in these two last cases by another white space).
My code is the following:
#include <iostream>
#include <string>
using namespace std;
int main()
{ string s;
cout << "Enter text (enter \"$\" to stop input):\n";
getline(cin,s,'$');
string s2 = s, word;
int ini = 0, last, count_word = 0;
int count_1 = 0, count_2 = 0, count_3 = 0;
cout << "\nThe words found in the text, with its frequencies, are the following:\n";
for (ini; ini < s.length(); )
{ // we look for the next word in the string (at the end of each iteration
// ini is incremented in a quantity previous_word.length()
last = ini;
cout << "1: " << ++count_1 << endl;
while(true)
{ if (s[last] == ' ') break;
if (s[last] == '\n') break;
if (s[last] == ',') break;
if (s[last] == '.') break;
if (last > s.length()-1 ) break;
++last;
cout << "2: " << ++count_2 << endl;
}
--last; // last gives the position within s of the last letter of the current word
// now me create the word itself
word = s.substr(ini,last-ini+1); //because last-ini is word.length()-1
int found = s2.find(word);
while( found != s2.length() ) // the loop goes at least once
++count_word;
s2.erase(0,found+word.length()); // we erase the part of s2 where we have already looked
found = s2.find(word);
cout << "3: " << ++count_3 << endl;
cout << "\t["<<word<<"]: " << count_word;
++last;
s2 = s;
s2.erase(0,ini + word.length()); // we do this so that in the next iteration we don't look for
// the new word where we know it won't be.
if (s[last] == ' ' || s[last] == '\n') ini = last + 1;
if (s[last] == ',' || s[last] == '.') ini = last + 2;
count_word = 0;
}
}
When I ran the program nothing was sshown on screen, so I figured out that one of the loops must had been stuck (that is why I defined the variables count_1,2 and 3, to know if this was so).
However, after correctly counting the number of iterations for the fist word to be found, nothing else is printed and all I see is the command prompt (I mean the tiny white bar) and I cannot even stop the program by using ctrl z.
This is a very complicated method for a very simple problem. You can just use a stringstream to extract each word seperated by a white space. You then just take the extracted word and increment the word counter using a std::map<std::string, int>.
My take on this:
#include <iostream>
#include <map>
#include <string>
#include <sstream>
int main() {
std::map<std::string, int> word_to_count;
std::string in;
std::getline(std::cin, in);
std::stringstream s(in);
std::string temp_word;
while (s >> temp_word) {
word_to_count[temp_word]++;
}
for (const auto& x : word_to_count) {
std::cout << x.first << ": " << x.second << std::endl;
}
return 0;
}
input
hello world hello world test
Output
hello: 2
test: 1
world: 2
Keep in mind this is just one of many possible solutions, so just take this as inspiration :).
I have prompted the user to input a string in the main function in my program and store it userString, and want to display how many words there are.
This is the function I intend to call from main:
int countWords(string d) {
string words = " ";
for (int e = 0; e < d.length(); e++) {
if (isspace(d[e])) {
cout << "The string " << words << "word(s). ";
}
}
return words;
}
I read somewhere that the function should actually count the number of white spaces (which is why I used isspace()), and not the words themselves.
How do I go about counting the number of words there are in the string and displaying it in the same function? I am having trouble figuring it out and I'm getting errors.
I also cannot use library functions.
Expected output:
The string "2020" has one word.
The string "Hello guys" has two words.
If you don't want to use boost, a simple for loop will do.
#include <cctype>
...
for(int i = 0; i < toParse.length(); i++){
if (isblank(toParse[i])){
//start new word
}
else if (toParse[i] == '.'){
//start new sentence
}
else if (isalphanum(toParse[i])){
//add to your current word
}
}
edit: you can just increment an integer where you see the //start new word.
try boost::split(), which will put words into a vector
Also, if you want to count something in a range satisfying some condition, you can think in std::count_if
Example:
int countWords(std::string d)
{
int w = std::count_if(d.begin(), d.end(), [](char ch) { return isspace(ch); });
std::cout << "The string \"" << d << "\" has " << w + 1 << " words." << '\n';
return w;
}
I am trying to create an anagram solver just using a very basic, procedural approach. I am finding out that I probably should have done this using classes, but now it is too late and my assignment is about due. Any suggestions on how to figure this out would be great!
Basically, this is what the algorithm should do:
Get all words in the dictionary; store them in a container
Get a word from the user; quit if appropriate
Get all permutations of the word that the user entered
Strip the word the user entered from the permutations
Strip all words in the permutation collection that aren't also in the dictionary I collected in part 1
Now for the last step, I must make sure that I don't display duplicate anagrams (i.e. anagrams which contain the same letter, such as "loop"). I cannot seem to get this check to work, which is noted below with under the TODO comment block.
Any suggestions would be awesome!!
#include <iostream>
#include <fstream>
#include <string>
//
// Change size below to accomodate more anagrams and dictionary words
//
#define MAX_ANGM_SIZE 4096
#define MAX_WORD_SIZE 1048576
using namespace std;
//
// Determines whether anagram is valid or not; will not display word
// which user entered or words not contained in dictionary
//
bool isValidAnagram(string word, string userWord,
string dictionary[], unsigned int listIdx)
{
for(unsigned int idx = 0; idx < listIdx; ++idx)
{
if(word == userWord)
return false;
else if (word == dictionary[idx])
return true;
}
return false;
}
//
// Determines whether user's word is contained in the dictionary
// or not
//
bool isValidWord(string word, string dictionary[],
unsigned int listIdx)
{
for(unsigned int idx = 0; idx < listIdx; ++idx)
{
if(word == dictionary[idx])
return true;
}
return false;
}
//
// TODO:This function should test for duplicate anagrams and return
// true if duplicates are found.
//
bool isRepeated(string anagrams[], unsigned int anaIdx)
{
for(unsigned int idx = anaIdx; idx != 0; --idx)
{
if(anagrams[idx] == anagrams[anaIdx])
return true;
else
return false;
}
return false;
}
//
// Only display elements in array which aren't blank and don't
// display duplicate anagrams; notify user if no anagrams
// were found.
//
void displayAnagrams(string anagrams[], unsigned int next)
{
int flag = 0;
for (unsigned int idx = 0; idx < next; ++idx)
{
if((anagrams[idx] != "") || (!(isRepeated(anagrams, idx))))
{
if(idx == 1)
cout << " Anagrams: ";
if(idx > 0)
flag = 1;
cout << anagrams[idx] << " ";
}
else
continue;
}
if(flag == 0)
cout << " no anagrams found" << endl;
}
static void swap(char &c1, char &c2)
{
char temp = c1;
c1 = c2;
c2 = temp;
}
//
// Pass in word to be altered, the userWord for comparison, the array to store
// anagrams, the dictionary for comparison, the count for the number of anagrams
// and the count for number of dictionary words
//
static void permute(string word, string userWord, int k, string anagrams[],
string dictionary[], unsigned int &next, unsigned int listIdx)
{
if(k == word.length()-1)
{
if(isValidAnagram(word, userWord, dictionary, listIdx))
anagrams[next] = word;
++next;
}
else
{
for(int idx = k; idx < word.length(); ++idx)
{
swap(word[k], word[idx]);
permute(word, userWord, k+1, anagrams, dictionary, next, listIdx);
}
}
}
//
// Create container to store anagrams, validate user's word in dictionary, get all
// of the anagrams, then display all valid anagrams
//
void getAnagrams(string word, string dictionary[], unsigned int listIdx)
{
string anagrams[MAX_ANGM_SIZE];
unsigned int next = 0;
if(isValidWord(word, dictionary, listIdx))
{
permute(word, word, 0, anagrams, dictionary, next, listIdx);
}
else
{
cerr << " \"" << word << "\"" << " is not a valid word" << endl;
return;
}
displayAnagrams(anagrams, next);
}
//
// Read in dictionary file, store contents of file in a list, prompt
// the user to type in words to generate anagrams
//
int main()
{
string file;
string word;
string quit = "quit";
string dictionary[MAX_WORD_SIZE];
unsigned int idx = 0;
cout << "Enter a dictionary file: ";
cin >> file;
cout << "Reading file \"" << file << "\"" << endl;
cout << endl;
ifstream inFile(file.c_str());
if(!(inFile.is_open()))
{
cerr << "Can't open file \"" << file << "\""
<< endl;
exit(EXIT_FAILURE);
}
while(!inFile.eof())
{
inFile >> dictionary[idx];
++idx;
}
inFile.close();
while(true)
{
cout << "Enter a word: ";
cin >> word;
if(word == quit) break;
getAnagrams(word, dictionary, idx);
cout << endl;
}
return 0;
}
You may want to rethink your step (3). If the user enters a 12-letter word you have 479,001,600 permutations of it which will probably be impractical to assemble all at once (and if that's not, then a 16-letter word will be...).
Instead, try thinking about how you could store the words and look up potential anagrams in a way that doesn't require you to do that.
Edit: I get that ability to solve largeish words may not be your biggest concern at this point, but it might actually make your fourth and fifth steps easier if you do them by assembling the set of valid words rather than starting with all possibilities and removing all the ones that don't match. 'Removing' an item from an array is a bit awkward since you have to shuffle all the following items up to fill in the gap (this is exactly the kind of thing that STL manages for you).
Better algorithm : don't store your word, but store a tupple containing (your word, sorted letters). Moreover, you sort that big storage by the second key (hint, you could use a sqlite database to do the work for you and use an index (can't be unique!))
E.g. to store
"Florent", "Abraham","Zoe"
you would store in memory
("aaabhmr", "abraham"),("eflnort","florent"),("eoz","zoe")
When you got your word from your user, you just use same "sorting letter inside word" algorithm.
Then you look for that pattern in your storage, and you find all anagrams very quickly (log(size of dictionary)) as it's sorted. Of course, original words are the second elements of your tuple.
You can do that using classes, standard structures, a database, up to you to choose the easiest implementation (and the one fitting your requirements)
What is the best STL to use for this task? I've been using Map,
and I couldn't get it to work. I'm not sure how I am supposed to check the number of same words that occur in the sentence for example:
I love him, I love her, he love her.
So I want the program to prompt the user to enter an integer, lets say i enter 3, the output will be love as the same word occurs 3 times in the sentence. But what method to use if I want to do a program like this?
Currently my program prompts for the user to enter the word, and then it shall return how many time that word occurs, which for word love, is 3. but now i want it the other way round. Can it be done? Using which STL will be better?
I assume you use a map to store the number of occurrences.
Well,you first have to understand this,since you are using a map,the key is unique while the stored data may not be unique.
Consider a map, x
with contents
x["I"]=3
x["Love"]=3
x["C"]=5
There is unique a mapping from the key to the value,and not the other way round,if you want this one to one mapping ,i would suggest a different data structure.If you want to use map,and still search for an element,using STL search function or your own.Or you can write your search function.
search().
map<string,int>::iterator ser;
cin>>check;
for(ser=x.begin();ser!=x.end();++ser)
{
if(ser->second==check)
{
cout<<"Word"<<ser->first<<endl;
break;
}
}
First build the mapping from word to count and then build the reverse multi-mapping from that. Finally, you can determine which words occur with a given frequency:
#include <algorithm>
#include <iostream>
#include <iterator>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <utility>
int main()
{
std::string str("I love him, I love her, he love her");
std::istringstream ss(str);
std::istream_iterator<std::string> begin(ss);
std::istream_iterator<std::string> end;
std::map<std::string, int> word_count;
std::for_each(begin, end, [&](const std::string& s)
{
++word_count[s];
});
std::multimap<int, std::string> count_words;
std::for_each(word_count.begin(), word_count.end(),
[&](const std::pair<std::string, int>& p)
{
count_words.insert(std::make_pair(p.second, p.first));
});
auto its = count_words.equal_range(3);
std::for_each(its.first, its.second,
[](const std::pair<int, std::string>& p)
{
std::cout << p.second << std::endl;
});
}
/******************************************************************
Name : Paul Rodgers
Source : HW1.CPP
Compiler : Visual C++ .NET
Action : Program will read in from standard input and determine the
frequency of word lengths found in input. An appropriate
table is also displayed. Maximum word length is 15 characters
words greater then 15 are counted as length 15.
Average word length also displayed.
Note : Words include hyphenated and ones with apostrophes. Words with
apostrophes, i.e. Jim's, will count the apostrophe as part of the
word length. Hyphen is counted if word on same line, else not.
Also an int array is used to hold the number of words with
length associated with matching subscript, with subscript 0
not being used. So subscript 1 corresponds to word length of 1,
subscript 2 to word length of 2 and so on.
------------------------------------------------------------------------*/
#include <iostream>
#include <ctype.h>
#include <iomanip>
using namespace std;
int NextWordLength(void); // function prototypes
void DisplayFrequencyTable(const int Words[]);
const int WORD_LENGTH = 16; // global constant for array
void main()
{
int WordLength; // actual length of word 0 to X
int NumOfWords[WORD_LENGTH] = {0}; // array holds # of lengths of words
WordLength = NextWordLength();
while (WordLength) // continue to loop until no word, i.e. 0
{ // increment length counter
(WordLength <= 14) ? (++NumOfWords[WordLength]) : (++NumOfWords[15]);
WordLength = NextWordLength();
}
DisplayFrequencyTable(NumOfWords);
}
/********************** NextWordLength ********************************
Action : Will determine the length of the next word. Hyphenated words and
words with apostrophes are counted as one word accordingly
Parameters : none
Returns : the length of word, 0 if none, i.e. end of file
-----------------------------------------------------------------------*/
int NextWordLength(void)
{
char Ch;
int EndOfWord = 0, //tells when we have read in one word
LengthOfWord = 0;
Ch = cin.get(); // get first character
while (!cin.eof() && !EndOfWord)
{
while (isspace(Ch) || ispunct(Ch)) // Skips leading white spaces
Ch = cin.get(); // and leading punctation marks
if (isalnum(Ch)) // if character is a letter or number
++LengthOfWord; // then increment word length
Ch = cin.get(); // get next character
if ((Ch == '-') && (cin.peek() == '\n')) //check for hyphenated word over two lines
{
Ch = cin.get(); // don't count hyphen and remove the newline char
Ch = cin.get(); // get next character then on next line
}
if ((Ch == '-') && (isalpha(cin.peek()))) //check for hyphenated word in one line
{
++LengthOfWord; // count the hyphen as part of word
Ch = cin.get(); // get next character
}
if ((Ch == '\'') && (isalpha(cin.peek()))) // check for apostrophe in word
{
++LengthOfWord; // count apostrophe in word length
Ch = cin.get(); // and get next letter
}
if (isspace(Ch) || ispunct(Ch) || cin.eof()) // is it end of word
EndOfWord++;
}
return LengthOfWord;
}
/*********************** DisplayFrequencyTable **************************
Action : Will display the frequency of length of words along with the
average word length
Parameters
IN : Pointer to array holding the frequency of the lengths
Returns : Nothing
Precondition: for loop does not go beyond WORD_LENGTH
------------------------------------------------------------------------*/
void DisplayFrequencyTable(const int Words[])
{
int TotalWords = 0, TotalLength = 0;
cout << "\nWord Length Frequency\n";
cout << "------------ ----------\n";
for (int i = 1; i <= WORD_LENGTH-1; i++)
{
cout << setw(4) << i << setw(18) << Words[i] << endl;
TotalLength += (i*Words[i]);
TotalWords += Words[i];
}
cout << "\nAverage word length is ";
if (TotalLength)
cout << float(TotalLength)/TotalWords << endl;
else
cout << 0 << endl;
}
#include<iostream>
#include<string>
#include<vector>
#include<cstddef>
#include<map>
using std::cout;
using std::cin;
using std::string;
using std::endl;
using std::vector;
using std::map;
int main() {
cout << "Please enter a string: " << endl;
string str;
getline(cin, str, '\n');
size_t str_len = str.size();
cout << endl << endl;
size_t i = 0, j = 0;
bool pop = false;
map<string, int> myMap;
for (size_t k = 0; k < str_len-1; k++) {
if (((k == 0) && isalpha(str[0])) || (!(isalpha(str[k-1])) && isalpha(str[k])))
i = k;
if ( isalpha(str[k]) && !(isalpha(str[k+1])) ) {
j = k;
pop = true;
}
if ( (k == str_len-2) && isalpha(str[k+1]) ) {
j = k+1;
pop = true;
}
if ( (i <= j) && pop ) {
string tmp = str.substr(i, j-i+1);
cout << tmp << '\t';
myMap[tmp]++;
pop = false;
}
}
cout << endl << endl;
map<string, int>::iterator itr, end = myMap.end();
for (itr = myMap.begin(); itr != end; itr++)
cout << itr->first << "\t - - - - - \t" << itr->second << endl;
cout << endl;
return 0;
}