How to Convert Python Lists to Dictionaries and Sort by Weight Efficiently

Introduction

Hello, I'm PiPi. I recently saw an interesting question on Zhihu and decided to share the solution.

Implementation

The problem involves converting a list to a dictionary and then sorting the dictionary by the numeric values of its entries, which may be expressed in tons ('t') or kilograms ('kg').

Basic conversion example:

d = ['熊', '1.3t', '海鸥', '88kg', '彭', '99kg', '凤', '0.68t']
rs = {d[i]: d[i+1] for i in range(0, len(d), 2)}
print(rs)

To sort the dictionary by weight, one approach is:

dict1 = {'熊': '1.3t', '海鸥': '88kg', '彭': '99kg', '凤': '0.68t'}
list1 = []
for i, j in dict1.items():
    list1.append([i, j])
print(list1)

def getWeight(weight):
    if weight[-1] == 't':
        return float(weight[:-1]) * 1000
    elif weight[-2:] == 'kg':
        return float(weight[:-2])
    return 0
animals = sorted(list1, key=lambda animal: getWeight(animal[1]))
print(animals)

More concise solutions were contributed by other users, for example:

d = ['熊', '1.3t', '海鸥', '88kg', '彭', '99kg', '凤', '0.88t']
d1 = dict(sorted(zip(d[::2], d[1::2]), key=lambda x: float(x[1].replace('t',''))*1000 if 't' in x[1] else float(x[1].replace('kg',''))))
print(d1)

For nested list structures, the conversion and sorting can be performed as follows:

animals = [['熊', '1.3t'], ['海鸥', '88kg'], ['彭', '99kg'], ['凤', '0.68t']]
animals_dict = dict(animals)
print(animals_dict)
d1 = list(animals_dict.items())
d1.sort(key=lambda x: float(x[1][:-1])*1000 if '.' in x[1] else int(x[1][:-2]))
print(dict(d1))

Finally, converting a dictionary back to a list of pairs:

dict1 = {'熊': '1.3t', '海鸥': '88kg', '彭': '99kg', '凤': '0.68t'}
list1 = []
for i, j in dict1.items():
    list1.append([i, j])
print(list1)

Conclusion

The article provides a comprehensive overview of converting Python lists to dictionaries, sorting the resulting mapping by weight, and handling various extensions such as nested lists and reverse conversion, with multiple code snippets to help readers solve the problem.