如何使用 python 按字典的值对文件进行排序

Question

    file=sorted(file)

词典数据（姓名、平均分数）较早收集并存储在文件中。此代码然后重新打开文件并将其中的数据存储在变量（文件）中。我需要文件中的这些数据按值排序，所以如果文件中的数据是：詹姆斯：5 蒂姆：8 它将进行排序，以便蒂姆排在首位，因为他的平均得分更高。我将如何去做这件事？字典也在那里，因此 name 和 score 在排序时保持在一起

Answer 1

import collections

dict = {}

# Assume file format is "name:score" for each line.
with open("classa2.txt", "r") as f:
    for line in f:
        line = line.split(':')
        dict[line[0]] = line[1]

# Sort by value (score)
od = collections.OrderedDict(sorted(dict.items(), key=lambda t: t[1]))

# Write new results
with open("classa2.txt", "w") as f:
    for key in od:
        f.write(key + ':' + str(od[key]) + "\n")
        print key + ":" + str(od[key])

Answer 2

数据文件看起来像

tim: ["4.0", "2.9", "3.4"]
james: ["3.8", "3.8"]
clara: ["2.6", "3.5", "3.2"]

在这种情况下你需要像

这样的东西

from ast import literal_eval

def sort_key(line):
    # split "name: scores" into ["name", " scores"]
    name, score_str = line.split(":")

    # convert " [\"1.9\", \"3.8\"]" into ["1.9", "3.8"]
    scores = literal_eval(score_str.strip())

    # convert ["1.9", "3.8"] into [1.9, 3.8]
    scores = [float(f) for f in scores]

    # calculate average score
    avg = sum(scores) / len(scores)
    return avg

那你可以

# read data from file
with open("classa2.txt") as inf:
    lines = [line.rstrip() for line in inf]

# sort highest scores first
lines.sort(key=sort_key, reverse=True)

print("\n".join(lines))

这导致

james: ["3.8", "3.8"]
tim: ["4.0", "2.9", "3.4"]
clara: ["2.6", "3.5", "3.2"]

如何使用 python 按字典的值对文件进行排序

how to sort files by the value of a dictionary using python

python

sorting

dictionary