Question

我得到了项目列表及其得分。如何返回每个子文件夹中包含最高项目的列表

来自

[('../dir_a/1.png', 5.14), 
 ('../dir_a/2.png', 5.15), 
 ('../dir_b/3.png', 4.19), 
 ('../dir_b/4.png', 3.81)]

收件人

[('../dir_a/2.png', 5.15), 
 ('../dir_b/3.png', 4.19)]

Answer 1

import os

result = {}
lst = [('../dir_a/1.png', 5.14), ('../dir_a/2.png', 5.15), ('../dir_b/3.png', 4.19), ('../dir_b/4.png', 3.81)]

for p in lst:
    base_dir =  os.path.basename(os.path.dirname(p[0]))     #Use base dir as key
    if base_dir not in result:
        result[base_dir] = ("", 0)
    if p[1] > result[base_dir][1]:                          #check score
        result[base_dir] = (p)
print(result.values())                                      #Get values

输出：

[('../dir_a/2.png', 5.15), ('../dir_b/3.png', 4.19)]

Answer 2

这是使用熊猫的方法：

# Create example data frame
df = pd.DataFrame([('../dir_a/1.png', 5.14), 
                   ('../dir_a/2.png', 5.15), 
                   ('../dir_b/3.png', 4.19), 
                   ('../dir_b/4.png', 3.81)], columns = ['path', 'score'])

# Split the file path by '/' and expand into columns with original data frame
df = pd.concat([df.path.str.split('/', expand=True), df], axis=1)

# Group the rows based on the directory name (column 1) and find the max score
df.groupby(1)['score'].max().reset_index()

       1  score
0  dir_a   5.15
1  dir_b   4.19

然后，您可以根据需要将值转换回列表。

Answer 3

如果只需要最大值，则可以将目录放在字典中并从中删除，然后保留最大值。

from pathlib import PurePath

max_dict = {}
for path, val in lst:
    parent = PurePath(path).parent
    max_dict[parent] = max(max_dict.get(parent, float('-inf')), val)

Answer 4

这里是单线：

a = [('../dir_a/1.png', 5.14), ('../dir_a/2.png', 5.15), ('../dir_b/3.png', 4.19), ('../dir_b/4.png', 3.81)]

f = [(next((x[0] for x in a if x[1]==max([v[1] for v in a if v[0].split('/')[1]==k])),k),max([v[1] for v in a if v[0].split('/')[1]==k])) for k in set([kk[0].split('/')[1] for kk in a])]

输出：

[('../dir_a/2.png', 5.15), ('../dir_b/3.png', 4.19)]

有一个项目和分数列表。如何返回每个子文件夹中具有最高项目的列表

4 个答案: