min() function should return the smallest of the numbers given. Then why <pre class="prettyprint"><code>min({0},{1}) </code></pre> returns {0}, while <pre class="prettyprint"><code>min({1},{0}) </code></pre> returns {1}?

In python sets are compared based on whether or not one is a subset of the other. None is a subset of the other, so <code><</code> gives <code>False</code> in all cases, so the first one is returned. So <code>{1}<{0}</code> gives <code>False</code> just like <code>{2}<{1}</code> gives <code>False</code> and <code>{2}<{3,4,5}</code> gives False. However <code>{1,2}<{1,3,2}</code> gives <code>True</code>. This means there is no total order defined on sets.

Why min({1},{0}) returns {1} [duplicate]

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python

min

min() function should return the smallest of the numbers given. Then why

min({0},{1})

returns {0}, while

min({1},{0})

returns {1}?

865

asked Dec 03 '22 22:12

Adam Ježek

1 Answers

In python sets are compared based on whether or not one is a subset of the other.

None is a subset of the other, so < gives False in all cases, so the first one is returned.

So {1}<{0} gives False just like {2}<{1} gives False and {2}<{3,4,5} gives False. However {1,2}<{1,3,2} gives True. This means there is no total order defined on sets.

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