Question

我正在尝试找出列表中不存在的某些元素。

说，我有l1由正则表达式（某种模式）组成

l1 = ["file.log","sample.log","abc_log_(\d+)_(\d+)_(\d+)_test-analysis.log","abc_(\d+)_(\d+)_(\d+)_sample-analysis.log"]


# l1 consisting regexes are of standard set of files.

test1 = ["file.log","sample.log","abc_log_123_12_12_test-analysis.log","abc_145_20_20_sample-analysis.log"]

说test1是测试列表，将其与l1进行比较以检查是否生成了所有文件。 In this case all files are present.

类似地，test2 = ["file.log","abc_145_20_20_sample-analysis.log"] 将test2与l1进行比较时，应通知未生成sample.log和以test-analysis.log结尾的文件。

    How can this be done with minimum complexity ?

请找到以下代码

import re

l1 = ["file.log","sample.log","abc_log_(\d+)_(\d+)_(\d+)_test-analysis.log","abc_(\d+)_(\d+)_(\d+)_sample-analysis.log"]
test1 = ["file.log","sample.log","abc_log_123_12_12_test-analysis.log","abc_145_20_20_sample-analysis.log"]
#test1 = ["file.log","abc_145_20_20_sample-analysis.log"]

for i in l1:
    flag = ""
    tmp = []
    for j in test1:
        if re.match("^"+str(i)+"$",j):
            flag = "yes"
            tmp.append(True)
            print "File {} present".format(i)
            break
    if flag != "yes":
        print "File not present : {}".format(i)
        tmp.append(False)

而且，请提出是否有更好的方法/方法。

Answer 1

如果test1列表中的顺序无关紧要，则可以将any()函数与list comprehension一起使用，从而避免使用临时变量。

for i in l1:
    if any([re.match("^"+str(i)+"$",j) for j in test1]):
        print "File {} present".format(i)
    else:
        print "File not present : {}".format(i)

如何找到与正则表达式列表不匹配的元素？

1 个答案: