Question

在下面的代码中，我将应用程序名称分配给进程ID（硬编码），例如：在下面的代码中，1468是进程ID，outlook是进程名称。在硬编码值中输入了每个pid的cpu相关信息。这里我想用进程名替换进程id，并且应该得到相应进程名输出的cpu相关信息，如值字典中所示。

预期输出

outlook: [
    233.625,
    336.15625,
    43791.0,
    168.21875,
    94.859375
], 
chrome: [
    3713.625,
    3396.15625,
    36791.0,
    628.21875,
    48.859375
],
command prompt:[
    3713.625,
    396.15625,
    3791.0,
    628.21875,
    48.859375
]

实际输出：

3796
12972
1468
process name for this pid outlook
{3796: [3713.625, 396.15625, 3791.0, 628.21875, 48.859375], 12972: [3713.625, 3396.15625, 36791.0, 628.21875, 48.859375], 1468: [233.625, 336.15625, 43791.0, 168.21875, 94.859375]}

代码：

import os, platform, subprocess, re 
import sys
import psutil
import itertools 
import json

#if __name__ == "__main__" :
if True:
    dict = {'1468 ': 'outlook', '12972':'chrome', '3796': 'command prompt'}

    values ={
                1468 : [
                    233.625,
                    336.15625,
                    43791.0,
                    168.21875,
                    94.859375
                ], 
                12972: [
                    3713.625,
                    3396.15625,
                    36791.0,
                    628.21875,
                    48.859375
                ],
                3796:[
                    3713.625,
                    396.15625,
                    3791.0,
                    628.21875,
                    48.859375
                ]
            }
class pretty:
    def printf(self, json_object):
        print(json.dumps(json_object, sort_keys=True, indent=4, separators=
(',', ': ')))

    def comparision(self):
        for iter in values.keys():
            print (iter) 
        for proc in psutil.process_iter():
            if iter == proc.pid:
               #print(iter)
                print("process name for this pid" +" "+proc.name())
                if iter == proc.pid:
                    print(values)

pretty.comparision("")

请帮我解决这个问题，我是python的新手

Answer 1

避免使用dict和values作为变量名称，因为它们是dictionary个关键字，并将它们用作变量会导致意外结果：

使用：

dict1 = {'1468 ': 'outlook', '12972':'chrome', '3796': 'command prompt'}
values1 ={
    1468 : [
        233.625,
        336.15625,
        43791.0,
        168.21875,
        94.859375
    ], 
    12972: [
        3713.625,
        3396.15625,
        36791.0,
        628.21875,
        48.859375
    ],
    3796:[
        3713.625,
        396.15625,
        3791.0,
        628.21875,
        48.859375
    ]
}

>>> res = {}
>>> for key, value in dict1.items():
...    res[dict1[key]] = values1[int(key)]
... 
>>> res
{'chrome': [3713.625, 3396.15625, 36791.0, 628.21875, 48.859375], 'outlook': [233.625, 336.15625, 43791.0, 168.21875, 94.859375], 'command prompt': [3713.625, 396.15625, 3791.0, 628.21875, 48.859375]}

Answer 2

你不能这样做吗

for key in dict:
    print (dict[key], ":", values[int(key)])

如果需要的话，我可以添加更多解释。

Answer 3

dict = {'1468': 'outlook', '12972':'chrome', '3796': 'command prompt'}
values ={1468 : [233.625,336.15625,43791.0,168.21875,94.859375],
12972: [3713.625,3396.15625,36791.0,628.21875,48.859375],
3796:[3713.625,396.15625,3791.0,628.21875,48.859375]}
new_dict={}

for k in values.keys():
    new_dict[dict[str(k)]] = str(values[k])

print new_dict # this would give you a dictionary

打印字典内容的问题

3 个答案: