从字典创建列表，或只是简单地对它进行排序

Question

我有以下代码：

import os 
import pprint 

file_path = input("Please, enter the path to the file: ") 
if os.path.exists(file_path): 
    worker_dict = {} 
    k = 1 
    for line in open(file_path,'r'): 
     split_line = line.split() 
     worker = 'worker{}'.format(k) 
     worker_name = '{}_{}'.format(worker, 'name') 
     worker_yob = '{}_{}'.format(worker, 'yob') 
     worker_job = '{}_{}'.format(worker, 'job') 
     worker_salary = '{}_{}'.format(worker, 'salary') 
     worker_dict[worker_name] = ' '.join(split_line[0:2]) 
     worker_dict[worker_yob] = ' '.join(split_line[2:3]) 
     worker_dict[worker_job] = ' '.join(split_line[3:4]) 
     worker_dict[worker_salary] = ' '.join(split_line[4:5]) 
     k += 1 
else: 
    print('Error: Invalid file path') 

文件：

John Snow 1967 CEO 3400$ 
Adam Brown 1954 engineer 1200$ 

从worker_dict输出：

{ 
'worker1_job': 'CEO', 
'worker1_name': 'John Snow', 
'worker1_salary': '3400$', 
'worker1_yob': '1967', 
'worker2_job': 'engineer', 
'worker2_name': 'Adam Brown', 
'worker2_salary': '1200$', 
'worker2_yob': '1954', 
} 

而且我想按工人姓名对数据进行排序，然后按工资排序。所以我的想法是创建一个单独的名单与工资和工人名称进行排序。但是我填写它有问题，也许有一种更优雅的方式来解决我的问题？

Answer 1

import os 
import pprint 

file_path = input("Please, enter the path to the file: ") 
if os.path.exists(file_path): 
    worker_dict = {} 
    k = 1 
    with open(file_path,'r') as file: 
     content=file.read().splitlines() 
    res=[] 
    for i in content: 
     val = i.split() 
     name = [" ".join([val[0],val[1]]),]#concatenate first name and last name 
     i=name+val[2:] #prepend name 
     res.append(i) #append modified value to new list 
    res.sort(key=lambda x: x[3])#sort by salary 
    print res 
    res.sort(key=lambda x: x[0])#sort by name 
    print res 

输出：

[['Adam Brown', '1954', 'engineer', '1200$'], ['John Snow', '1967', 'CEO', '3400$']] 
[['Adam Brown', '1954', 'engineer', '1200$'], ['John Snow', '1967', 'CEO', '3400$']] 

Answer 2

d = { 
'worker1_job': 'CEO', 
'worker1_name': 'John Snow', 
'worker1_salary': '3400$', 
'worker1_yob': '1967', 
'worker2_job': 'engineer', 
'worker2_name': 'Adam Brown', 
'worker2_salary': '1200$', 
'worker2_yob': '1954', 
} 

from itertools import zip_longest 

#re-group: 
def grouper(iterable, n, fillvalue=None): 
     "Collect data into fixed-length chunks or blocks" 
     # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx" 
     args = [iter(iterable)] * n 
     return zip_longest(*args, fillvalue=fillvalue) 

#re-order: 
res = [] 
for group in list(grouper(d.values(), 4)): 
      reorder = [1,2,0,3] 
      res.append([ group[i] for i in reorder]) 

#sort: 
res.sort(key=lambda x: (x[1], x[2])) 

输出：

[['Adam Brown', '1200$', 'engineer', '1954'], 
['John Snow', '3400$', 'CEO', '1967']] 

石斑鱼被定义并在itertools说明。我已经按照每个工作人员的记录对您的字典进行了分组，并将其作为列表的重新排序列表返回。作为名单，我按名称和薪水对他们进行分类。这是解决方案是模块化的：它明显地分组，重新排序和排序。

Answer 3

我建议以不同的格式存储工作人员，例如.csv，然后您可以使用csv.DictReader并将其放入词典列表中（这也可以让您使用更多词汇的工作，名称等） “古墓丽影”）。

请注意，您必须出生和工资的年转换为整数或漂浮到他们正确地排序，否则他们会得到字典顺序排序在一个真实的世界词典（书），因为它们都是字符串，如：

>>> sorted(['100', '11', '1001']) 
['100', '1001', '11'] 

要排序列表的排列，您可以使用operator.itemgetter作为sorted的关键参数，而不是lambda函数，只需将所需的键传递给itemgetter即可。

k变量没用，因为它只是列表的len。

.csv文件：

"name","year of birth","job","salary" 
John Snow,1967,CEO,3400$ 
Adam Brown,1954,engineer,1200$ 
Lara Croft,1984,tomb raider,5600$ 

的.py文件：

import os 
import csv 
from operator import itemgetter 
from pprint import pprint 


file_path = input('Please, enter the path to the file: ') 
if os.path.exists(file_path): 
    with open(file_path, 'r', newline='') as f: 
     worker_list = list(csv.DictReader(f)) 
     for worker in worker_list: 
      worker['salary'] = int(worker['salary'].strip('$')) 
      worker['year of birth'] = int(worker['year of birth']) 

    pprint(worker_list) 
    pprint(sorted(worker_list, key=itemgetter('name'))) 
    pprint(sorted(worker_list, key=itemgetter('salary'))) 
    pprint(sorted(worker_list, key=itemgetter('year of birth'))) 

你仍然需要一些错误处理，如果INT转换失败，或只是让程序崩溃。