检查两个哈希值是否具有相同的键集

Question

什么是检查两个哈希值h1并且h2具有相同的密钥集而无视订单的最有效方法？是否可以比我发布的答案更快或更简洁,效率更高？

Answer 1

好吧,让我们打破所有的生活和便携性规则.MRI的C API开始发挥作用.

/* Name this file superhash.c. An appropriate Makefile is attached below. */
#include <ruby/ruby.h>

static int key_is_in_other(VALUE key, VALUE val, VALUE data) {
  struct st_table *other = ((struct st_table**) data)[0];
  if (st_lookup(other, key, 0)) {
    return ST_CONTINUE;
  } else {
    int *failed = ((int**) data)[1];
    *failed = 1;
    return ST_STOP;
  }
}

static VALUE hash_size(VALUE hash) {
  if (!RHASH(hash)->ntbl)
    return INT2FIX(0);
  return INT2FIX(RHASH(hash)->ntbl->num_entries);
}

static VALUE same_keys(VALUE self, VALUE other) {
  if (CLASS_OF(other) != rb_cHash)
    rb_raise(rb_eArgError, "argument needs to be a hash");
  if (hash_size(self) != hash_size(other))
    return Qfalse;
  if (!RHASH(other)->ntbl && !RHASH(other)->ntbl)
    return Qtrue;
  int failed = 0;
  void *data[2] = { RHASH(other)->ntbl, &failed };
  rb_hash_foreach(self, key_is_in_other, (VALUE) data);
  return failed ? Qfalse : Qtrue;
}

void Init_superhash(void) {
  rb_define_method(rb_cHash, "same_keys?", same_keys, 1);
}

这是一个Makefile.

CFLAGS=-std=c99 -O2 -Wall -fPIC $(shell pkg-config ruby-1.9 --cflags)
LDFLAGS=-Wl,-O1,--as-needed $(shell pkg-config ruby-1.9 --libs)
superhash.so: superhash.o
    $(LINK.c) -shared $^ -o $@

人工,综合和简单的基准显示了以下内容.

require 'superhash'
require 'benchmark'
n = 100_000
h1 = h2 = {a:5, b:8, c:1, d:9}
Benchmark.bm do |b|
  # freemasonjson's state of the art.
  b.report { n.times { h1.size == h2.size and h1.keys.all? { |key| !!h2[key] }}}
  # This solution
  b.report { n.times { h1.same_keys? h2} }
end
#       user     system      total        real
#   0.310000   0.000000   0.310000 (  0.312249)
#   0.050000   0.000000   0.050000 (  0.051807)

Answer 2

结合freemasonjson和sawa的想法:

h1.size == h2.size and (h1.keys - h2.keys).empty?

Answer 3

尝试:

# Check that both hash have the same number of entries first before anything
if h1.size == h2.size
    # breaks from iteration and returns 'false' as soon as there is a mismatched key
    # otherwise returns true
    h1.keys.all?{ |key| !!h2[key] }
end

枚举所有#？

更糟糕的情况是,你只需要遍历一次密钥.