检查迭代期间字典值中是否存在子字符串

Question

以下代码运行良好，但任何人都可以指导如何简化它。 newrelicid 是一个字典，其中包含“数字”作为键和“某些字符串”作为值（字符串可以包含子字符串，如 prod、test、local 等 ) 基本上我试图忽略包含来自列表['dev','staging','qat','uat','local','eu']的子字符串的值（ j ）@

样本newrelicid值

{1: 'Service PROD', 2: 'service', 3: 'guess-service (Production)', 4: 'check-service (Dev)', 5: 'analytics-service (Staging)' }

循环：

for i,j in newrelicid.items():
  if 'staging' not in j.lower() and not 'qat' in j.lower() and not 'uat' in j.lower() and not 'local' in j.lower() and not 'eu' in j.lower() and not 'dev' in j.lower():
    print(j)    

Answer 1

您可以使用any() 有效地模拟嵌套迭代：

>>> newrelicid = {
...     1: 'Service PROD',
...     2: 'service',
...     3: 'guess-service (Production)',
...     4: 'check-service (Dev)',
...     5: 'analytics-service (Staging)'
... }
>>> blacklist = {'dev','staging','qat','uat','local','eu'}

>>> {v for v in newrelicid.values()
...     if not any(i in v.casefold() for i in blacklist)}
{'Service PROD', 'guess-service (Production)', 'service'}

这会产生一个set[str] 的传递值。

Answer 2

你可以使用正则表达式：

import re

pattern = r'staging|local|uat|qat|dev|eu' # one pattern for all keywords

newrelicid = {1: 'Service PROD', 2: 'service', 3: 'guess-service (Production)',
              4: 'check-service (Dev)', 5: 'analytics-service (Staging)' }

for j in newrelicid.values():
    if re.search(pattern,j.lower()): # match with lowercase
        print(j)

check-service (Dev)
analytics-service (Staging)

如果你需要的是整个单词而不仅仅是子字符串，你可以改进模式：

pattern = r'\b(staging|local|uat|qat|dev|eu)\b'

Answer 3

您似乎想应用过滤器。

顺便说一句，由于您只查看字典中的字符串，请使用 dict.values()

l = ['staging','qat', 'uat', 'local', 'eu', 'dev']

list(filter(lambda x : not any(e in x.lower() for e in l),  newrelicid.values()))

['Service PROD', 'service', 'guess-service (Production)']