英文:
combining dictionaries to keep same values along with different values
问题
You can combine the dictionaries in Python by iterating through them and checking for unequal values. Here's the code to achieve the expected output:
x = {'a': 1, 'b': 2, 'c': 3, 'd': 2}
y = {'a': 1, 'b': 3, 'c': 4, 'd': 2}
z = {}
for key in x:
if x[key] != y[key]:
z[key] = f"{x[key]} to {y[key]}"
else:
z[key] = x[key]
print(z)
This code will give you the desired output:
{'a': 1, 'b': '2 to 3', 'c': '3 to 4', 'd': 2}
英文:
I have two dictionaries with the values.
x = {'a': 1, 'b': 2, 'c': 3, 'd':2}
y = {'a': 1, 'b': 3, 'c': 4, 'd':2}
How could I combine the values in python where these are not equal and if equal keep the same?
Expected output:
z = {'a': 1, 'b': 2 to 3, 'c': 3 to 4, 'd':2}
答案1
得分: 0
Here are the translations of the code sections you provided:
- 假设两个字典中的键相同，使用字典推导式和条件语句：
z = {k: v - y[k] if v != y[k] else v for k, v in x.items()}
输出: {'a': 1, 'b': -1, 'c': -1, 'd': 2}
- 对于元组：
z = {k: (v, y[k]) if v!=y[k] else v for k, v in x.items()}
输出: {'a': 1, 'b': (2, 3), 'c': (3, 4), 'd': 2}
- 对于字符串：
z = {k: f'{v} to {y[k]}' if v!=y[k] else v for k, v in x.items()}
输出: {'a': 1, 'b': '2 to 3', 'c': '3 to 4', 'd': 2}
- 用于任意数量的字典的通用方法：
dicts = [x, y]
out = {}
for d in dicts:
for k, v in d.items():
out.setdefault(k, set()).add(v)
out = {k: ', '.join(map(str, v)) for k, v in out.items()}
输出: {'a': '1', 'b': '2, 3', 'c': '3, 4', 'd': '2'}
英文:
Assuming the keys are identical in the two dictionaries, use a dictionary comprehension with a conditional:
z = {k: v - y[k] if v != y[k] else v for k, v in x.items()}
Output: {'a': 1, 'b': -1, 'c': -1, 'd': 2}
edited answer
For a tuple:
z = {k: (v, y[k]) if v!=y[k] else v for k, v in x.items()}
Output: {'a': 1, 'b': (2, 3), 'c': (3, 4), 'd': 2}
And a string:
z = {k: f'{v} to {y[k]}' if v!=y[k] else v for k, v in x.items()}
Output: {'a': 1, 'b': '2 to 3', 'c': '3 to 4', 'd': 2}
generic approach for an arbitrary number of dictionaries
dicts = [x, y]
out = {}
for d in dicts:
for k, v in d.items():
out.setdefault(k, set()).add(v)
out = {k: ', '.join(map(str, v)) for k, v in out.items()}
Output: {'a': '1', 'b': '2, 3', 'c': '3, 4', 'd': '2'}
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