How to avoid errors when dictionary is empty?

Question

I have a df where I need to flag a column as '1' if the 'row in my .apply() matches an item my dictionary. However, if my dictionary is empty or doesn't include the same 'Key' as the row which is in my .apply() at that instance, the script fails. How can I progress through this hickup?

df =  pd.DataFrame({'Key': ['10003', '10003', '10003', '10003', '10003','10003','10034'], 
               'Num1': [12,13,30,12,13,13,16],
               'Num2': [121,122,122,124,125,126,127],
              'admit': [2015019, 20150124, 20150206,20150211,20150220,20150407,20150211],
          'discharge': [20150123, 20150202, 20150211,20150220,20150304,20150410,20150211]})
df['admit'] = pd.to_datetime(df['admit'], format='%Y%m%d')
df['discharge'] = pd.to_datetime(df['discharge'], format='%Y%m%d')
#df=df.head(5)

script:

d2 = df[df['discharge'].isin(range(30,40))].groupby('Key')['discharge'].apply(set).to_dict()

def find(x):
    match2 = x['admit'] in d2[x['Key']]
    return match2

df['flag'] = df.apply(find, axis=1).astype(int)

In particular, I need to flag a column where the admit date of one row is equal to the discharge date of another AND the row with the matching discharge date has a value in Num1 between 30-40. This script works as expected if you reduce the df to just 5 rows df=df.head(5). But when there are rows where their 'Key' is not in the dictionary, the script returns an error. I am wondering if adding all 'key' and blank dates for the dictionary would make this work?

KeyError: ('10034', 'occurred at index 6')

I want to use dictionaries to perform the task as described above because the rest of my function has similar conditions as this (those were more simple). And the code above works as it should on a small sample, but my dictionary experience is low and this is stumping me. Sorry if this is a simple and stupid question.

final df:

    Key    Num1 Num2    admit   discharge   flag
0   10003   12  121 2015-01-09  2015-01-23  0
1   10003   13  122 2015-01-24  2015-02-02  0
2   10003   30  122 2015-02-06  2015-02-11  0
3   10003   12  124 2015-02-11  2015-02-20  1
4   10003   13  125 2015-02-20  2015-03-04  0
5   10003   13  126 2015-04-07  2015-04-10  0
6   10034   16  127 2015-02-11  2015-02-11  0

Answer 1

You can use the dict.get and return a empty list.

Ex:

def find(x):
    match2 = x['admit'] in d2.get(x['Key'], [])
    return match2

Answer 2

Use a try / except clause to catch KeyError and specify what to return in this situation:

def find(x):
    try:
        return x['admit'] in d2[x['Key']]
    except KeyError:
        return False