如何使用循环判断字典所有元素均大于另一字典对应元素？

Alright, let's break this down. We need to use a while loop to verify that every value in a person's dictionary is greater than the corresponding value in a requirements dictionary—only returning True if all conditions are satisfied. Here's a straightforward approach with your example data:

Step-by-Step Implementation #

First, let's define our example dictionaries and write a function to handle the check:

# Example dictionaries
requirements = {"happiness": 9, "money": 100}
person1 = {"happiness": 15, "money": 3200}
person2 = {"happiness": 4, "money": 5000}

def meets_all_requirements(reqs, person):
    # Extract the list of keys (we assume both dicts share identical keys)
    requirement_keys = list(reqs.keys())
    current_index = 0
    all_requirements_met = True
    
    # Loop through each key using a while loop
    while current_index < len(requirement_keys):
        key = requirement_keys[current_index]
        # Check if the person's value falls below the requirement
        if person[key] < reqs[key]:
            all_requirements_met = False
            # Optional: You can break early here since we already know the result is False
            # break
        current_index += 1
    
    return all_requirements_met

# Test the function with our examples
print(meets_all_requirements(requirements, person1))  # Output: True
print(meets_all_requirements(requirements, person2))  # Output: False

How It Works #

• Initialize Variables: We start by converting the requirements' keys into a list so we can iterate through them with an index. We set current_index to 0 (to start at the first key) and all_requirements_met to True—we assume all requirements are satisfied until we find one that isn't.
• While Loop Check: For each key, we compare the person's value against the requirement. If any value is less than the requirement, we flip all_requirements_met to False.
• Iterate Through Keys: We increment the index each time to move to the next key until we've checked every requirement.
• Return Result: Finally, we return the boolean indicating whether all requirements were met.

Extra Robustness (Optional) #

If you want to handle cases where the dictionaries might have different keys, you can add a quick check at the start of the function:

if reqs.keys() != person.keys():
    raise ValueError("Requirements and person dictionaries must have identical keys")

内容的提问来源于stack exchange，提问作者by James