def is_odd(n):
return n % 2 == 1
t = list(filter(is_odd, [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]))
print(t)
#注意到filter()函数返回的是一个Iterator，也就是一个惰性序列，所以要强迫filter()完成计算结果，需要用list()函数获得所有结果并返回list。
def not_empty(s):
return s and s.strip()
g = ['A', '', 'B', None, 'C', ' ']
print(g)
t = list(filter(not_empty, g))
print(t)s = ' rain '
print(s.strip())t = 'www.example.com'.strip('comowle.z')
print(t)def _odd_iterator():
n = 1
while(True):
n = n + 2
yield n
def _not_divisible(n):
return lambda x : x % n > 0
def primes():
it = _odd_iterator()
yield 2
while(True):
n = next(it)
yield n
it = filter(_not_divisible(n), it)for n in primes():
if n < 1000:
print(n)
else:
break
上面是filter
下面是sort
t = sorted([1, 5, 6, -19, 54])
print(t)t = sorted([1, 5, 6, 7, -19, 54], key = abs)
print(t)#忽略大小写对首字母的顺序进行排序，实际上是根据ascii码进行排序
t = sorted(['rain', 'SUN', 'rainbow', 'Thinker'], key = str.lower)
print(t)
#要实现反向排序，不必更改key函数，可传入第三个参数，reverse = Truet = sorted(['rain', 'SUN', 'rainbow', 'Thinker'], key = str.lower, reverse = True)
print(t)L = [('Bob', 75), ('Adam', 92), ('Bart', 66), ('Lisa', 88)]#按名字排序
def by_name(t):
#其中t[0]是t的名字，比如Bob，t[1]是t的分数，比如75
return t[0]
L2 = sorted(L, key=by_name)print('按照成绩排序', L2)
#按照成绩排序
def by_score(t):
return t[1]
L2= sorted(L, key = by_score, reverse = True)
print('按照分数排序', L2)
