正则表达式的第一个方法
(?!...)
如果匹配。。。与下一个不匹配。这是一个消极的前瞻性Assert。

import re

regex_filter = 'buy_train_20(?!18)\d*-\d*-\d*_\d*_0'

expr1 = 'buy_train_2018-10-16_617901_0'
m = re.search(regex_filter, expr1)
print(m)

# None

# (if None not do not try to print)

expr2 = 'buy_train_2020-01-08_001221_0'
m = re.search(regex_filter, expr2)
print(m)
print(m.group(0))

# <_sre.SRE_Match object; span=(0, 29), match='buy_train_2020-01-08_001221_0'>

# buy_train_2020-01-08_001221_0

具有筛选器本机函数的第二种方法：
但您不一定需要使用regex进行过滤，只需使用本机过滤函数，如下所示：

paths = ['buy_train_2020-01-08_001221_0',
         'buy_train_2020-05-02_067341_0',
         'buy_train_2020-07-26_011901_0',
         'buy_train_2018-10-16_617901_0',
         'buy_train_2018-12-19_492111_0']

prefix = 'buy_train_2018'

def function(path):
    if path[:len(prefix)] == prefix:
        return False
    else:
        return True

results = filter(function, paths)

for res in results:
    print (res)

# buy_train_2020-01-08_001221_0

# buy_train_2020-05-02_067341_0

# buy_train_2020-07-26_011901_0

import re

regex_filter = 'buy_train_20(?!18)\d*-\d*-\d*_\d*_0'

expr1 = 'buy_train_2018-10-16_617901_0'
m = re.search(regex_filter, expr1)
print(m)

# None

# (if None not do not try to print)

expr2 = 'buy_train_2020-01-08_001221_0'`enter code here`
m = re.search(regex_filter, expr2)
print(m)
print(m.group(0))

# <_sre.SRE_Match object; span=(0, 29), match='buy_train_2020-01-08_001221_0'>

# buy_train_2020-01-08_001221_0