如何在字典列表中查找值并返回特定密钥

Question

我有一个gpg密钥列表。每个都在自己的字典中。我想通过一个值'foo@foo.bar'来搜索这个字典列表s，并在同一个字典中返回关键字d ['keyid']的值。下面是我有：

>>> import gnupg 
>>> gpg = gnupg.GPG() 
>>> email = 'foo@foo.bar' 
>>> keyIn = gpg.gen_key_input(name_real = 'works' , name_email = 
>>> email,key_type='RSA',key_length=1024) 
>>> key = gpg.gen_key(keyIn) 
>>> klist = gpg.list_keys() 
>>> print klist 
[{'dummy': u'', 'keyid': u'CF7BBCC34CCC28A0', 'expires': u'1420347600', 'subkeys': [], 'length': u'1024', 'ownertrust': u'u', 'algo': u'1', 'fingerprint': u'5D917845BF81E47403265380CF7BBCC34CCC28A0', 'date': u'1388889639', 'trust': u'u', 'type': u'pub', 'uids': [u'works <foo@foo.bar>']}, {'dummy': u'', 'keyid': u'865F4A95D4999F17', 'expires': u'1420347600', 'subkeys': [], 'length': u'1024', 'ownertrust': u'u', 'algo': u'1', 'fingerprint': u'C081E6552B027E2DA4143852865F4A95D4999F17', 'date': u'1388890408', 'trust': u'u', 'type': u'pub', 'uids': [u'works <foo@foo.bar>']}, {'dummy': u'', 'keyid': u'1F6A4AEA477EFBD6', 'expires': u'1420347600', 'subkeys': [], 'length': u'1024', 'ownertrust': u'u', 'algo': u'1', 'fingerprint': u'B306139A3D740CECA71D9F281F6A4AEA477EFBD6', 'date': u'1388890595', 'trust': u'u', 'type': u'pub', 'uids': [u'works <foo@foo.bar>']}] 

现在，这里是我试过的功能：

def idfind(klist): 
    global email 
    for k in klist: 
     if email in k.items(): 
      return k 

def idfind(l): 
    global email 
    for d in l: 
     if any(d[email]): 
      return d['keyid'] 

def idfind(l, val): 
    for d in l: 
     for v in d[k]: 
      if val in v: 
       return d['keyid'] 

def idfind(lst,val): 
    for d in lst: 
     if isinstance(d,dict): 
      for k,v in d.items(): 
       if val in k: 
        return d['keyid'] 

def idfind(lst,val): 
    for d in lst: 
     if isinstance(d,dict): 
      for k,v in d.items(): 
       if val == v: 
        return d['keyid'] 

的功能全部或者返回None或错误。我如何得到这个工作。为什么我失败了。我哪里错了？

Answer 1

您可以使用列表中理解和any功能，这样

key = 'foo@foo.bar' 
print [d["keyid"] for d in klist if any(key in uid for uid in d["uids"])] 

输出

[u'CF7BBCC34CCC28A0', u'865F4A95D4999F17', u'1F6A4AEA477EFBD6'] 

Answer 2

试试这个，可能工作：

def idfind(klist): 
    for d1 in a: 
     for e in d1['uids']: 
      if email in e: 
       print d1['keyid'] 

输出：

CF7BBCC34CCC28A0 
865F4A95D4999F17 
1F6A4AEA477EFBD6