How to replace values from specific keys in listed dictionaries? But replace should happen with a conditional (i.e. if they have 2 digits in the beginning). Preferably with list/dictionary comprehensions.
Example:
a_list = [{'key1': 'A picture 1', 'key2': 'location 1', 'key3': '20 title1'}, {'key1': 'A picture 2', 'key2': 'location 2', 'key3': '10 title2'}]
I want to remove the first two digits from the values of key3.
I would do that using a dictionary comprehension within a list comprehension.
The dictionary comprehension rebuilds the dictionary, but alters the value if:
key3 ANDAlteration is taking the right part of the leading space.
a_list = [{'key1': 'A picture 1', 'key2': 'location 1', 'key3': '20 title1'}, {'key1': 'A picture 2', 'key2': 'location 2', 'key3': '10 title2'}]
a_new_list = [{k:v.partition(" ")[2] if (k=="key3" and v[:2].isdigit()) else v for k,v in d.items()} for d in a_list]
print(a_new_list)
result:
[{'key1': 'A picture 1', 'key2': 'location 1', 'key3': 'title1'}, {'key1': 'A picture 2', 'key2': 'location 2', 'key3': 'title2'}]
Note: Will also remove the first digits if there are more than 2 digits. The condition is broad and should be tuned to your needs.
EDIT: a pre Python 2.7 compliant alternative (dict comprehensions when did not exist):
a_new_list = [dict((k,v.partition(" ")[2] if (k=="key3" and v[:2].isdigit()) else v) for k,v in d.items()) for d in a_list]
There may be a way to do this with comprehensions, but in my opinion it is better to be a little more verbose to make the code's intention more clear:
import re
for d in a_list:
for k in d:
d[k] = re.sub(r'^\d\d', '', d[k]) if k == 'key3' else d[k]
So I combined Jean-François Fabre's and Tom Lynch's answers, slightly altered regex patterned and came up with this:
import re
a_new_list = [{k: re.sub(r'^\d{2} ', '', d[k]) if k == "key3" else v for k, v in d.items()} for d in a_list]
print a_new_list
