擅长:python、mysql、java
<p>我希望这会有帮助</p>
<pre><code>dict_a= {1: 488, 2: 336, 3: 315, 4: 291, 5: 275}
a = set(dict_a)
dict_b={2: 0, 3: 33, 1: 61, 5: 90, 15: 58}
b = set( dict_b)
dict_c= {1: 1.15, 9: 0, 2: 0.11, 15: 0.86, 19: 0.008, 20: 1834}
c = set( dict_c )
a_intersect_b = a & b
a_intersect_c = a & c
b_intersect_c = b & c
a_interset_b_intersect_c = a_intersect_b & c
total_intersect = {}
for id in a_interset_b_intersect_c:
total_intersect[id] = { dict_a[id] , dict_b[id] , dict_c[id] }
print total_intersect
a_b_only_intersect = {}
for id in a_intersect_b:
a_b_only_intersect[id] = { dict_a[id] , dict_b[id] }
print a_b_only_intersect
b_c_only_intersect = {}
for id in b_intersect_c:
b_c_only_intersect[id] = { dict_b[id] , dict_c[id] }
print b_c_only_intersect
a_c_only_intersect = {}
for id in a_intersect_c:
a_c_only_intersect[id] = { dict_a[id] , dict_c[id] }
print a_c_only_intersect
</code></pre>
<p>类似地,你可以用集合的“差”来找出a,b和c中的剩余部分。你知道吗</p>