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How to sort a list of strings by frequency?


By : xym2010
Date : July 29 2020, 09:00 AM
hope this fix your issue Your algorithm is unecessarily quadratic time. The following is linear
code :
from collections import Counter
resl2 = [k for k,v in Counter(resl).items() if v > 500]
resl2 = [(k,v) for k,v in Counter(resl).items() if v > 500]
resl2.sort(key=lambda kv: kv[1])
resl2 = [k for k,v in resl2]


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How to sort Strings by frequency then by first occurrence


By : Blake Arney
Date : March 29 2020, 07:55 AM
wish helps you Hello maybe this question already answered somewhere else, but I could find it. I have text file which includes , Try this
code :
  public static void main(String[] args) throws IOException {

    Scanner sc = new Scanner(new File("file.txt"));
    List<String> lines = new ArrayList<String>();
    while (sc.hasNextLine()){
        lines.add(sc.nextLine());
    }


    String[] arr = lines.toArray(new String[0]);
    String text = Arrays.toString(arr);
    String test = text.replaceAll("\\p{P}","");

    List<String> list = Arrays.asList(test.split(" ")); 
    SortedSet<String> uniq = new TreeSet<String>(list);


    for (String w : uniq){
        System.out.printf("%n%d     %s",Collections.frequency(list, w), w);

    }
}

Frequency Sort Array Of Objects in JS and if frequency Matches sort on basis of object property


By : Swapnil Madhukar Wam
Date : March 29 2020, 07:55 AM
like below fixes the issue I have Array of objects , Try this one using reduce and then sort.
Explaination
code :
var arr = [
    {"X" : {
        "price" : "5"
      }
    },
    {"Y" : {
        "price" : "3"
      }
    },
    {"Y" : {
        "price" : "3"
      }
    },
    {"Z" : {
        "price" : "4"
      }
    },
    {"Q" : {
        "price" : "2"
      }
    },
    {"X" : {
        "price" : "5"
      }
    },
    {"Z" : {
        "price" : "4"
      }
    },
    {"X" : {
        "price" : "5"
      }
    }
];

var frequency = arr.reduce(function (accumulatorObject, currentValue) { 
  var propName = Object.keys(currentValue)[0];
  var elementInArray = accumulatorObject.find((element) => Object.keys(element)[0] === propName);
  if (elementInArray) {
    elementInArray[propName].count++;
  }
  else {
    var newObject = {};
	newObject[propName]  = {};
	newObject[propName].count = 1;
	newObject[propName].price = +currentValue[propName].price;
	accumulatorObject.push(newObject);
 }
  return accumulatorObject;
}, []); //  // Accumulator starts with Empty array. 



frequency.sort(function(first,second){ 
    var diff =  second[Object.keys(second)].count - first[Object.keys(first)].count;
	if( diff === 0) {
		return second[Object.keys(second)].price - first[Object.keys(first)].price;
	}
return diff;});
console.log(frequency);

Sort a (nested) list of strings on frequency of occurrence


By : BmanUltima
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You can't use Counter directly because there are some list mixed in with string, therefore flatten those first:
code :
data = ['Marie', 'Lucas', 'Patsy', 'Julie', 'Peter', 'Emma', 'Kevin', 'Tobias', 'Suzy', ['Marie', 'Lucas'], ['Lucas', 'Marie'], ['Lucas', 'Patsy'], ['Patsy', 'Lucas'], ['Emma', 'Lucas'], ['Lucas', 'Emma'], ['Emma', 'Kevin'], ['Kevin', 'Emma'], ['Peter', 'Emma'], ['Emma', 'Peter'], ['Peter', 'Lucas'], ['Lucas', 'Peter'], ['Peter', 'Julie'], ['Julie', 'Peter'], ['Suzy', 'Tobias'], ['Tobias', 'Suzy']]
from collections import Counter
c = Counter()
for d in data:
    if isinstance(d,str):
        c[d] += 1
    elif isinstance(d,list):
        for n in d:
            c[n] += 1
c.most_common()
[('Lucas', 9),
 ('Peter', 7),
 ('Emma', 7),
 ('Tobias', 3),
 ('Patsy', 3),
 ('Marie', 3),
 ('Suzy', 3),
 ('Kevin', 3),
 ('Julie', 3)]

Sort a list by frequency of strings in it


By : Sangeetha R
Date : March 29 2020, 07:55 AM
Does that help List is , First, count the occurrence of string and then sort by using Map

Count frequency of words in a list and sort by frequency


By : onsitesigns
Date : March 29 2020, 07:55 AM
like below fixes the issue I am using Python 3.3 , use this
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