Difference between revisions of "Sets"

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(Created page with "<table border="1" cellpadding="5" cellspacing="0" rules="none" summary=""> <tr> <td> <table border="0" cellpadding="5" cellspacing="0" summary=""> <tr> <td val...")
 
 
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<p>For example,</p>
 
<p>For example,</p>
  
<apll>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'miasma'§'sis'<br />
+
<apll><pre>
mama<br />
+
      'miasma'§'sis'
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'miasma'§⍦'sis' ⍝ Using the Multiset form<br />
+
mama
mamas<br />
+
      'miasma'§⍦'sis' ⍝ Using the Multiset form
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'immiss'⊆'mississippi'<br />
+
mamas
1<br />
+
      'immiss'⊆'mississippi'
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'immiss'⊆⍦'mississippi' ⍝ Using the Multiset form<br />
+
1
0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ⍝ because the # m's doesn't match<br />
+
      'immiss'⊆⍦'mississippi' ⍝ Using the Multiset form
</apll>
+
0                             ⍝ because the # m's doesn't match
 +
</pre></apll>

Latest revision as of 23:21, 15 April 2018

Z←L§R returns a vector consisting of the elements of L that are not in R.
L is a scalar or one-element vector.
R is a scalar or one-element vector.
Z is the vector result equivalent to (L~R),R~L.


Z←L⊆R returns a Boolean scalar indicating whether or not L is a subset of R.
L is a scalar or one-element vector.
R is a scalar or one-element vector.
Z is the Boolean scalar result equivalent to ∧/L∊R as well as R⊇L.


Z←L⊇R returns a Boolean scalar indicating whether or not L is a superset of R.
L is a scalar or one-element vector.
R is a scalar or one-element vector.
Z is the Boolean scalar result equivalent to ∧/R∊L as well as R⊆L.


These functions behave differently when invoked via the Multiset Operator which takes into account multiplicities.

For example,

      'miasma'§'sis'
mama
      'miasma'§⍦'sis' ⍝ Using the Multiset form
mamas
      'immiss'⊆'mississippi'
1
      'immiss'⊆⍦'mississippi' ⍝ Using the Multiset form
0                             ⍝ because the # m's doesn't match