How to read from dictionary?

Question

Hi I have a dictionary as follows: static Dictionary ^> ^ language_string_table; where i have 17 string's. For each of 17 string's there is an associated list of 1500 string's. Now what i want to do is: compare strings in all the lists at each and every index. EX: 1st list 1st element should be compared with all the list's 1st element. Similarly for all the indexes in all the lists.

Vulpes · Answer

OK I think I've grasped it now: StreamWriter ^sw = gcnew StreamWriter(filePath); // look for empty strings for each(String ^key in language_string_table->Keys) { for(int v = 0; v { if (language_string_table[key][v] == ) sw->WriteLine(0x00); } } // look for common strings List ^common = gcnew List (); common->Add(); if(language_string_table->Values->Count > 1) { for each(String ^key in language_string_table->Keys) { for(int v = 0; v { bool foundDuplicate = false; String ^next = language_string_table[key][v]; // next value to compare with other values if (common->Contains(next)) continue; // already found this one for(int i = 0; i { for each(String ^k in language_string_table->Keys) { if (language_string_table[k][i] == next && (key != k || v != i)) { common->Add(next); foundDuplicate = true; break; // no point looking any further } } if (foundDuplicate) { int numBytes = 2 * next->Length; // all strings are unicode in .NET sw->WriteLine(String::Format({0}{1} numBytes GetHexString(next))); break; } } } } } // look for left over strings for each(String ^key in language_string_table->Keys) { for(int v = 0; v { String ^next = language_string_table[key][v]; if (!common->Contains(next)) { int numBytes = 2 * next->Length; // all strings are unicode in .NET sw->WriteLine(String::Format({0}{1} numBytes GetHexString(next))); } } } sw->Close(); // helper method String ^GetHexString(String ^s) { array ^sa = gcnew array (s->Length); int i = 0; for each(Char c in s) sa[i++] = String::Format(0x{0:X2} (int)c); return String::Join( sa); }

Vulpes · Answer

If you wanted to compare all the values in every list in the dictionary for equality you could do it like this: bool areEqual = true; if(language_string_table->Values->Count > 1) { // get the first value from the dictionary String ^ first; for each(String ^key in language_string_table->Keys) { first = language_string_table[key][0]; break; } for(int i = 0; i { for each(String ^key in language_string_table->Keys) { String ^value = language_string_table[key][i]; if (value != first) { areEqual = false; break; } } } } if(areEqual) { // do something } else { // do something else }

Vulpes · Answer

It depends on what you mean by 'compare' but you could read the strings at the same index in each list into a temporary array for further processing as follows: for(int i = 0; i { array ^temp = gcnew array (17); int j = 0; for each(String ^key in language_string_table->Keys) { temp[j++] = language_string_table[key][i]; } // do something with temp }

Karthik Agarwal · Answer

Hi Vulpes That's a perfect answer. Can you tell me an extension of the same question which i asked above? What should i do if had to compare all the elements in all the lists. Ex: just like comparing 1st element of the 1st list with all the elements of all the lists and should judge whether they are same or not? Thanks & Regards P.Karthik

Vulpes · Answer

I'm pleased to hear it worked Karthik as I could only test it on a small dataset so wasn't sure how it would scale to lists with 1575 members.

Karthik Agarwal · Answer

Great job Vulpes that's so nice of you. I must say that you didn't a brilliant job. I know you might have worked very hard for this. Because of weekend i couldn't try this out even though you replied on the same day. Today i worked out. It works with the small modifications i made on that in order to suit my application. Thank's a lot Vulpes. Really very much thanks. I mean it.

Karthik Agarwal · Answer

It has become too typical i guess. No if the string doesn't match with any other string it's called left over irrespective of index in any list. If the string matches with any other string in any list that is called as common string irrespective of index and list. From the example you stated 1: Common strings are: dog cat duck 2: left over strings are: lion tiger swan Now it will clear for i hope.

Vulpes · Answer

I'm sorry to press this but it's important to be clear as it affects the algorithm used. So to be regarded as common a string does have to appear in every list (regardless of index) but if there are duplicates of a common string in the same list they are not regarded as 'left over' strings? For example: List 1: dog cat dog duck duck List 2: dog cat lion tiger swan dog and cat are common as appear in both lists. The second instance of dog in the first list is also common not left over. duck is not common as it doesn't appear in list 2. It therefore needs two 'left over' entries.

Karthik Agarwal · Answer

No vulpes i don't wanna confuse you much but i still need to encounter each and every common string as i need to generate hexadecimal value for each and every occurrence of a common string also but in a different manner for common strings when compared to other two cases. Sorry for troubling you much.

Vulpes · Answer

Just a clarification on point 3. So if common strings can occur in the same list am I right in thinking that a common string doesn't have to appear in every list? To take an extreme example if a string only appeared twice in the same list but in no other list then it would still be regarded as common?

Karthik Agarwal · Answer

The answers are as follows: 1. I will write a for each loop to generate hexadecimal value after getting decimal value of the same. The for each loop automatically returns 0 as it doesn't contain any characters. Hence i will 0x00 to external file by appending 0x to the string. 2. Ya all the occurrences of empty strings needs to be written to the file 3. The common strings need not be in same index in all the lists to treat that as common strings. In can be any index and in any list even in the same list. 4. Only the hexadecimal value is to be written to the external file. And the format in which i want to write in external file is as follows 1. If it's an empty string i will write 0x00 for all empty strings 2. If it's a common string i will write as no.of bytes occupied by current string each character hex value separated by comma 3. and if it's neither empty string nor it didn't match with any other string no.of bytes occupied by current string each character hex value separated by comma I guess i answered everything you asked.

Vulpes · Answer

That raises some questions: 1. How can you generate a hexadecimal value for an empty string when by definition it has no characters? 2. Do all occurrences of empty strings need to be written to the file? 3. Do strings have to be in the same index position in each list to be regarded as 'common' or could they be in different index positions? 4. What exactly needs to be written to the file just the hexadecimal value of the characters (space delimited say) or the key and list index at which they are found?

Karthik Agarwal · Answer

Hi Vulpes This has become a pretty long discussion which i haven't expected. I thought that if i explain everything you might get confused hence i explained some part of my project. So issues are coming one after the other. Now what i intended to do completely in my project is as follows: I have 17 languages(Keys) and each 1575 list of strings(Values) associated for each language. What i need to do exactly with this is: 1. first i need to look for empty Values in all the lists and then if there are any empty strings i should generate hexadecimal value for the same. 2. Second i need to look for common strings i.e common Values in all the lists and generate hexadecimal value for the same. 3. For the left over strings in all the lists i should generate hexadecimal value in the same way. Note: in the 2nd point which i stated for all the common strings i should generate hexadecimal value only once. And i am using stream writer to write this hexadecimal values to output files. Last post of your's wasn't looking for common strings in the same language list rather it was looking only in other lists.

Vulpes · Answer

I've added an extra line (highlighted) to prevent some unnecessary processing.

Karthik Agarwal · Answer

OK I will check out and let you know the update.

Vulpes · Answer

OK this code should do it: if(language_string_table->Values->Count > 1) { for each(String ^key in language_string_table->Keys) { for(int v = 0; v { bool areEqual = true; String ^next = language_string_table[key][v]; // next value to compare with other lists for(int i = 0; i { for each(String ^k in language_string_table->Keys) { if (key == k) continue; // only compare with other lists String ^value = language_string_table[k][i]; if (value != next) { areEqual = false; break; } } if(!areEqual) break; // EDIT } if(areEqual) { // do something } else { // do something else } } } }

Karthik Agarwal · Answer

i want to do this for each and every value of each key.?

Vulpes · Answer

Do you want to do this for all the values of the first key only or for all the values of every key ?

Karthik Agarwal · Answer

Ya  Are you perhaps saying that you want to test the first element of the first list for equality with all the elements of all the other lists the second element of the first list for equality with the same set of elements and so on?  This is want i want to do....

Vulpes · Answer

I don't really follow that because either all the elements of all the lists are equal to each other or they're not. It doesn't matter which one you start with. Are you perhaps saying that you want to test the first element of the first list for equality with all the elements of all the other lists the second element of the first list for equality with the same set of elements and so on?

Karthik Agarwal · Answer

Hi Vulpes There is a small issue in the post which i accepted as answer. The issue is that the control doesn't come to this place more than once first = language_string_table[key][0]; hence string doesn't get updated. So all the time the only 1st list 1st element will be compared to all other elements in all other lists. What i was expecting is i should compare 1st list 1st element with all other elements and if it matches then do some task then i should take the 2nd element in the 1st list and compare with all other elements and if it matches then do the same task which i did previously similarly for all the elements in all the lists. Can you help me out.?

Karthik Agarwal · Answer

Hi Suthish Nair Ya i liked the way Vulpes answers all of my questions. Hence so many likes. Appreciation makes people work more efficiently what i feel.

Suthish Nair · Answer

21 Likes great...