内置函数filter

Posted 2020-09-20 IlyMxing

python3.6

class filter(object):
    """
    filter(function or None, iterable) --> filter object
    
    Return an iterator yielding those items of iterable for which function(item)
    is true. If function is None, return the items that are true.
    """

filter(func，iterator)

    func：自定义或匿名函数中所得值是布尔值，true将保留函数所取到的值，false则取反。

    iterator：可迭代对象。

例：

     过滤列表[‘text_test_text‘, ‘test_text_1‘, ‘text_test_2‘, ‘3_test_text‘, ‘test_test‘]

     只要含有text字符串及将其取出 or 取反。

s.rfind’text’+1

     Python3中 rfind() 返回字符串最后一次出现的位置，如果没有匹配项则返回-1。

     数字中0是false，0以上的整数都是true，所以s.rfind’text’后会有+1，没找到字符及-1+1=0.

# Filter

li = [‘text_test_text‘, ‘test_text_1‘, ‘text_test_2‘, ‘3_test_text‘, ‘test_test‘]

# 默认保留函数所取到的值
print(list(filter(lambda s: s.rfind(‘text‘) + 1, li)))
# 取反，下三个例子是一样的
print(list(filter(lambda s: not s.rfind(‘text‘) + 1, li)))

# Noe 自定义函数

l1 = [‘text_test_text‘, ‘test_text_1‘, ‘text_test_2‘, ‘3_test_text‘, ‘test_test‘]


def distinguish(l):
    nl = []
    for s in l:
        if s.rfind("text") + 1:
            nl.append(s)
    return nl


print(distinguish(l1))

# Two 自定义高阶函数

l2 = [‘text_test_text‘, ‘test_text_1‘, ‘text_test_2‘, ‘3_test_text‘, ‘test_test‘]


def f(s):
    return s.rfind(‘text‘) + 1


def distinguish(func, array):
    nl = []
    for s in array:
        if func(s):
            nl.append(s)
    return nl


print(distinguish(f, l2))

# Three 匿名函数

l3 = [‘text_test_text‘, ‘test_text_1‘, ‘text_test_2‘, ‘3_test_text‘, ‘test_test‘]


def distinguish(func, array):
    nl = []
    for s in array:
        if func(s):
            nl.append(s)
    return nl

print(distinguish(lambda s: s.rfind(‘text‘) + 1, l3))