如何用另一个字典值替换字典的值

Question

Hi I have two lists which contain some dictionary with their values

list1 = [{'val-1': 0, 'val-2': 0, 'val-3': 0}, {'val-1': 0, 'val-2': 0, 'val-3': 0}, {'val-4': 0, 
'val-5': 0, 'val-6': 0}]

list2 = [{'val-1': 90, 'val-4': 89, 'val-3': 99}, {'val-2': 88, 'val-6': 55, 'val-1':100}]

The desired output would be something like this

output: [{'val-1': 90, 'val-2': 88, 'val-3': 99}, {'val-1': 90, 'val-2': 88, 'val-3': 99}, {'val-4': 
89,'val-5': 0, 'val-6': 55}]

How could I make this replacement?

what I did form list2 I select each dictionary and compare and replace the values present in list1 all dictionary

but I didn't get the correct output

Answer 1

From the list2 you may build a unique dict that holds the transformation mappings, if you have more than once occurence of a key you have 2 choices

1. keep first occurence, don't overwrite
2. keep last, overwrite each time

For keepfirst

list2 = [{'val-1': 90, 'val-4': 89, 'val-3': 99}, {'val-2': 88, 'val-6': 55, 'val-1': 100}]

# 1. Keep first (in fact overwrite but do backward)
update_dict = {k: v for d in list2[::-1] for k, v in d.items()}
# 2. Keep last
update_dict = {k: v for d in list2 for k, v in d.items()}

Then rebuild the original list1 structure with list and dict comprehension, using the update_dict

result = [{k: update_dict.get(k, v) for k, v in subdict.items()}
          for subdict in list1]
print(result)

---

With 1. keep first

[{'val-1': 90, 'val-2': 88, 'val-3': 99}, {'val-1': 90, 'val-2': 88, 'val-3': 99}, {'val-4': 89, 'val-5': 0, 'val-6': 55}]

With 2. keep last (changes val-1 )

[{'val-1': 100, 'val-2': 88, 'val-3': 99}, {'val-1': 100, 'val-2': 88, 'val-3': 99}, {'val-4': 89, 'val-5': 0, 'val-6': 55}]