[英]How do you change a specific element of a set?
在这段代码中,我试图将每次循环的集合的值与参数中传递的值(在本例中为a)进行比较。 但是有趣的是,它显示了当我为每个循环使用a时,每个元素都是整数。 如何获得没有控制台错误的整数到整数比较?
def remove(s,a,b):
c=set()
c=s
for element in c:
element=int(element)
if(element<a or element>b):
c.discard(element)
return c
def main():
remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)
main()
输出:
if(element<=a or element>=b):
TypeError: '>=' not supported between instances of 'int' and 'set'
您重新分配本地变量b
:
def remove(s,a,b):
b=set() # now b is no longer the b you pass in, but an empty set
b=s # now it is the set s that you passed as an argument
# ...
if(... element > b): # so the comparison must fail: int > set ??
使用集合理解的简短实现:
def remove(s, a, b):
return {x for x in s if a <= x <= b}
>>> remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)
{9, 2, 3, 4}
如果要int进行int比较,则将b作为s的列表。
def remove(s,a,b):
b = list(s)
for element in s:
element=int(element)
if(element< a or element > b):
b.remove(element)
return b
def main():
remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)
main()
来吧,为什么我们不缩短代码呢?
尝试这个:
def remove(s, a, b):
return s.difference(filter(lambda x: not int(a) < int(x) < int(b), s))
def main():
new_set = remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)
# {2, 3, 4, 9}
print(new_set)
main()
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