CODEWITHSUNDEEP
python
Baraa Zaid
Baraa Zaid

Reputation: 434

How to return a value when you have KeyError Python?

I am trying to turn a python's keys into lists, however, some keys are empty and I get KeyError. To avoid the KeyError, I want it to return an empty string so I can easily append this to a dataframe. Also, I want an efficient/better way to run the code below, since I am retrieving a large amount of information, and implementing this process manually is very time consuming.

My code:

ages= []
names = []

for i in range(len(test)):
    
    try:   
        age= test[i]["age"]
        ages.append(age)

    except KeyError:
        age= ""
        ages.append(age)

    try:
        name = test[i]["name"]
        names.append(name)
        
    
    except KeyError:
        name = ""
        names.append(name)

I have many other data points from this dict that I want to retrieve such as weight, height, etc. and doing a try/except for all them can be tedious for code. Is there an efficient way to recreate this code.

Upvotes: 0

Views: 2418

Answers (3)

Pi Marillion
Pi Marillion

Reputation: 4674

There are three main ways.

Let's assume you have the simplified dict:

names = {123: 'Bob Roberts', 456: 'Alice Albertsons'}

And we're going to look up a name with ID 789, and we want to get John Doe when 789 isn't found in the names dictionary.

Method 1: Use the get method, which accepts a default value:

name_789 = names.get(789, 'John Doe')

Method 2: Use the setdefault method, which accepts a default value, and will also add that default as the new value in the dict if needed:

name_789 = names.setdefault(789, 'John Doe')

Method 3: Create the dictionary as a defaultdict instead:

names = collections.defaultdict((lambda: 'John Doe'), [
    (123, 'Bob Roberts'), (456, 'Alice Albertsons')
])

name_789 = names[789]

Note: Method 1 (get) is often really useful for nested dictionaries.

For example: outer.get(outer_key, {}).get(middle_key, {}).get(inner_key) will return outer[outer_key][middle_key][inner_key] if possible, or just None if any of the dicts needed are missing.

Upvotes: 2

user107511
user107511

Reputation: 822

You can use the defaultdict collection instead of a simple dictionary.

It creates a default value for a "missing" key.

Upvotes: 2

Tranbi
Tranbi

Reputation: 12721

You can use the get method of dictionaries and also loop directly through your list:

ages= []
names = []
for t in test:
    ages.append(t.get("age", "")
    names.append(t.get("name", "")

Upvotes: 3

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