Create Hashmap using string array as value and Enum as Key in Java 8 Lambda-Stream

Question

I am trying to achieve below logic using Lambda-Stream in Java 8. (See below sample method). I put down below logic in old fashion so that it will be easy to create solution instead of my version of Lambda - Stream setup which is more complex and wrong as not getting end result.

public static HashMap<EnumObj,String> getHashMapData(String…args){
     int i = 0;
     HashMap<EnumObj,String> hashObj = new HashMap<EnumObj,String>();
     if(args.length <= 6){
           for(String arg : args){
                if(i == 0){
                     hashObj.put(EnumObj.FIRSTNAME,args[i]);
                }else if(i == 1){
                     hashObj.put(EnumObj.LASTNAME,args[i]);
                }else if{……….
                ………….
                }else if(i == 4 && args.length < 5){
                      hashObj.put(EnumObj.COMPANY,args[i]);
                      hashObj.put(EnumObj.COMANYBOSS,args[i+1]);
                }else if(i == 5 && args.length < 6){
                      hashObj.put(EnumObj.COMANYBOSS,args[i]);
                }
           }  
     }
}

Please ignore any lopping logic of if / else if as time ago lost touch of such looping logic. While I am keep trying on my logic thought Got to get some idea if someone else has done successfully or have suggestion to achieve it with simplest manner.

So far I use two Stream to achieve with Collections.toMap option for creating new map. But it has Stream with in a stream and more complex.

If I found solution by my self then will post that answer otherwise will vote for best solution.

Kind of hard to tell what you're doing. In the above code, i is always 0, since you never modify it. If you add an i++ in the loop, then the conditions if (i == 4 && args.length < 5) and if (i == 5 && args.length < 6) will always be false. — ajb
– ajb, Commented Oct 30, 2015 at 5:55
In any case, what I think you want is to process two streams in parallel, where one comes from the array {EnumObj.FIRSTNAME, EnumObj.LASTNAME, ...}, and the other comes from args. However, a quick search seems to indicate that Java doesn't provide a built-in way to process two streams in parallel (see stackoverflow.com/questions/24059837/…). So I don't think you're going to get a good solution that uses streams. — ajb
– ajb, Commented Oct 30, 2015 at 5:59
That code looks horrible no matter which fashion it's written in. Instead of trying to convert it to a lambda stream for no good reason, you should try to fix your logic. It looks like you're trying to parse variable length data, but your solution looks incredibly brittle. — Kayaman
– Kayaman, Commented Oct 30, 2015 at 6:22
The first thing which came to my mind when reading the title, was, where does the requirement to use a HashMap come from, when it is known that the key is an enum. Is it because, you think every Map has to be a HashMap? What is ruling out using EnumMap? — Holger
– Holger, Commented Oct 30, 2015 at 12:00
Guys...I have converted my lambda to something more generic. I have mention in my comment that I have not such programming like using i and for loop since long time. I did this to put question in understand manner. Apologies if you feel brittle (which I agree it is). this is just for understanding of logic to achieve as end result. hope this explains my point of view — PreTri
– PreTri, Commented Oct 30, 2015 at 23:10

JB Nizet · Accepted Answer · 2015-10-30 06:49:01Z

1

As already said in the comments, this code is incorrect. I'll assume that you have a variable length array of arguments, and that each argument corresponds to an enum constant of EnumObj.

If that's the case, all you need is a simple for loop:

Map<EnumObj, String> result = new HashMap<>();
EnumObj[] enums = EnumObj.values();
for (int i = 0; i < args.length; i++) {
    result.put(enums[i], args[i]);
}

You could do the same with streams, but you wouldn't gain anything, IMO:

Map<EnumObj, String> result = new HashMap<>();
EnumObj[] enums = EnumObj.values();
IntStream.range(0, args.length)
         .forEach(i -> result.put(enums[i], args[i]));

answered Oct 30, 2015 at 6:49

JB Nizet

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3 Comments

Holger Over a year ago

You are assuming that the enum’s intrinsic order matches the order of the args elements. Unfortunately, there is nothing in the OP’s code/question proving it.

JB Nizet Over a year ago

Yes, indeed, those are my assumptions. I'll be happy to delete my answer if the OP comes back and makes it clear that this is not what he wants. Or he can just create an array/list of enum constants in the right order, and then use the above code.

PreTri Over a year ago

I just want to confirm my understanding that I am not gaining any new thing with Lambda. My original code base is similar to your first option (obvious reason can't put exact code base due to critical information in it). but its more near to my solution and understanding. Appreciated all for your help.

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