Erinevus lehekülje "JavaPython:Hulk" redaktsioonide vahel

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(Uus lehekülg: '{{JavaPython}}')
 
1. rida: 1. rida:
{{JavaPython}}
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{{JavaPython-sisukord}}
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== Näide ==
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{|
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!Java
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!Python
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|-
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|<syntaxhighlight lang="java" line="1" >
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//hashset
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HashSet<String> aSet = new HashSet<String>();
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aSet.add("aaaa");
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aSet.add("bbbb");
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aSet.add("cccc");
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aSet.add("dddd");
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//iterate over set
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Iterator<String> iterator = aSet.iterator();
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while (iterator.hasNext()) {
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System.out.print(iterator.next() + " ");
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}
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HashSet<String> bSet = new HashSet<String>();
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bSet.add("eeee");
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bSet.add("ffff");
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bSet.add("gggg");
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bSet.add("dddd");
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//check if bSet is a subset of aSet
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boolean b = aSet.containsAll(bSet);
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//union - transform aSet
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//into the union of aSet and bSet
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aSet.addAll(bSet);
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//intersection - transforms aSet
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//into the intersection of aSet and bSet
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aSet.retainAll(bSet);
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//difference - transforms aSet
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//into the (asymmetric) set difference
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// of aSet and bSet.
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aSet.removeAll(bSet);
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</syntaxhighlight>
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|<syntaxhighlight lang="python" line="2" >
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aSet = set()
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aSet = set("one") # a set containing three letters
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#set(['e', 'o', 'n'])
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aSet = set(['one', 'two', 'three'])
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#set(['three', 'two', 'one'])
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#a set containing three words
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#iterate over set
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for v in aSet:
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    print v
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bSet = set(['three','four', 'five'])
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#union
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cSet = aSet | bSet
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#set(['four', 'one', 'five', 'three', 'two'])
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#intersection
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dSet = aSet & bSet
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#find elements in aSet not bSet
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eSet = aSet.difference(bSet)
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#add element
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bSet.add("six")
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#set(['four', 'six', 'five', 'three'])
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</syntaxhighlight>
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|}

Redaktsioon: 30. jaanuar 2016, kell 07:45

Java vs Python

Näide

Java Python
<syntaxhighlight lang="java" line="1" >

//hashset HashSet<String> aSet = new HashSet<String>(); aSet.add("aaaa"); aSet.add("bbbb"); aSet.add("cccc"); aSet.add("dddd");

//iterate over set Iterator<String> iterator = aSet.iterator(); while (iterator.hasNext()) { System.out.print(iterator.next() + " "); }

HashSet<String> bSet = new HashSet<String>(); bSet.add("eeee"); bSet.add("ffff"); bSet.add("gggg"); bSet.add("dddd");

//check if bSet is a subset of aSet boolean b = aSet.containsAll(bSet);

//union - transform aSet //into the union of aSet and bSet aSet.addAll(bSet);

//intersection - transforms aSet //into the intersection of aSet and bSet aSet.retainAll(bSet);

//difference - transforms aSet //into the (asymmetric) set difference // of aSet and bSet. aSet.removeAll(bSet); </syntaxhighlight>

<syntaxhighlight lang="python" line="2" >

aSet = set() aSet = set("one") # a set containing three letters

  1. set(['e', 'o', 'n'])

aSet = set(['one', 'two', 'three'])

  1. set(['three', 'two', 'one'])
  2. a set containing three words
  1. iterate over set

for v in aSet:

   print v

bSet = set(['three','four', 'five'])

  1. union

cSet = aSet | bSet

  1. set(['four', 'one', 'five', 'three', 'two'])
  1. intersection

dSet = aSet & bSet

  1. find elements in aSet not bSet

eSet = aSet.difference(bSet)

  1. add element

bSet.add("six")

  1. set(['four', 'six', 'five', 'three'])

</syntaxhighlight>