Efficiently Iterating over dictionary list values by skipping missing values Python 3 -

i have pandas dataframe

import pandas pd  df=pd.dataframe({'location': [ 'ny', 'sf', 'ny', 'ny', 'sf', 'sf', 'tx', 'tx', 'tx', 'dc'],                  'class': ['h','l','h','l','l','h', 'h','l','l','m'],                  'address': ['12 silver','10 fak','12 silver','1 north','10 fak','2 fake', '1 red','1 dog','2 fake','1 white'],                  'score':['4','5','3','2','1','5','4','3','2','1',]}) 

and want add 2 tags stored in dictionaries. note second dictionary not include key 'a'

df['tag1'] ='' df['tag2'] =''  tagset1 = {'a':['ny|sf'],           'b':['dc'],           'c':['tx'],           } key in tagset1:     df.loc[df.location.str.contains(tagset1[key][0]) & (df.tag1 == ''),'tag1'] = key   tagset2= {'b':['h|m'],           'c':['l'],           } key in tagset2:     df.loc[df.class.str.contains(tagset2[key][0]) & (df.tag2 == ''),'tag2'] = key  print (df) 

if want combine both dictionaries make code more readable , efficient should fill in spot in newtagset['a'][1] '' or there way make iterator ignore or skip position newtagset['a'][1] when iterating on position in list?

newtagset = {'a':['ny|sf', '',],           'b':['dc','h|m',],           'c':['tx','l',],           }   key in newtagset:     df.loc[df.location.str.contains(newtagset[key][0]) & (df.tag1 == ''),'tag1'] = key  key in newtagset:     df.loc[df.class.str.contains(newtagset[key][1]) & (df.tag2 == ''),'tag2'] = key  print (df) 

most solutions found uses itertools skip multiple iterations in loop python way?

there nothing wrong simple continue.

for key, value in newtagset.items():    # found dict.items cleaner     if not value[1]:         continue     df.loc... 

a bit of off topic:

& (df.tag1 == '') redundant. useful if had coincidences in values, lead unpredictable behavior, since dict not ordered.