Question

是否有更好/更清洁/更短的方式获得与以下相同的输出？

import plistlib
pl = plistlib.readPlist('/Users/username/Documents/wifi1.plist')

n = len(pl)
count = 0
while (count < n):
    print('----------------')
    print(pl[count]['NOISE'])
    print(pl[count]['RSSI'])
    print(pl[count]['SSID_STR'])
    print(pl[count]['BSSID'])
    count += 1

我试过了：

for sub_dict in pl.values():
    print(sub_dict['NOISE'], sub_dict['RSSI'], sub_dict['SSID_STR'], sub_dict['BSSID'])

但我明白了：

Traceback (most recent call last):
  File "plistread.py", line 17, in <module>
    for sub_dict in pl.values():
AttributeError: 'list' object has no attribute 'values'

Answer 1

你只需要：

for sub_dict in pl:

由于pl是一个列表，因此遍历该列表将依次为您提供每个子字典。

一个简单的例子：

>>> l = [1,2,3,4]
>>> for x in l:
...   print x,
... 
1 2 3 4

Answer 2

您可以使用＆gt;＆gt;＆gt;进入＆lt;＆lt;＆lt;声明

for sub_dict in pl:

http://docs.python.org/tutorial/controlflow.html#for-statements

Answer 3

pl是一个列表，而不是字典。

所以，你应该通过以下方式迭代：

for sub_dict in pl:
    print(sub_dict['NOISE'], sub_dict['RSSI'], sub_dict['SSID_STR'], sub_dict['BSSID'])

Answer 4

如果我理解正确：

keys = ['NOISE', 'RSSI' ... rest of keys]
input = [{'NOISE':1, 'RSSI':2 ... rest of data}, {'NOISE':11, 'RSSI':22 ... rest of data} ... rest of data]

data = [[sub_dict[key] for key in keys] for sub_dict in input]

python嵌套字典循环

4 个答案: