从多个数据源在python中创建嵌套字典

Question

由于原始文章不清楚，所以将其完全改写。我想要做的是逐行解析一些数据并创建一个字典。我想有更好的方法来组织这些数据。我试图去解决这个问题的原始方式并没有解释几件事情，所以我提出了这个问题。我通过逐行循环服务策略输出来将接口，策略名称的数据放在一起，然后拉出队列，丢弃和无缓冲区丢弃。我遇到的问题是它没有考虑额外的策略，因此数据的原始传递被覆盖。

服务策略输出：

GigabitEthernet11/1 

Service-policy output: Gi11_1 

Counters last updated 7191104 seconds ago 

Class-map: class-default (match-any) 
    0 packets, 0 bytes 
    30 second offered rate 0000 bps, drop rate 0000 bps 
    Match: any 
    Queueing 
    queue limit 33025 packets 
    (queue depth/total drops/no-buffer drops) 0/0/0 
    (pkts output/bytes output) 0/0 
    shape (average) cir 500000000, bc 2000000, be 2000000 
    target shape rate 500000000 

    Service-policy : child 

    Counters last updated 7191104 seconds ago 

    Class-map: class-default (match-any) 
     0 packets, 0 bytes 
     30 second offered rate 0000 bps, drop rate 0000 bps 
     Match: any 
     Queueing 
     queue limit 33025 packets 
     (queue depth/total drops/no-buffer drops) 0/0/0 
     (pkts output/bytes output) 0/0 
     bandwidth remaining ratio 100 

for ints, int_strings in zip(int_names, int_output): 
    counts.setdefault(ints, {}) 

    for line in int_strings.splitlines(): 
     matchpolicy = re.search(r'(Service-policy.*)', line) 
     matchdrops = re.findall(r'total drops.*', line) 
     if matchpolicy: 
      spolicies = matchpolicy.group(0) 
      counts[ints]['Policy'] = spolicies 
     if matchdrops: 
      regx = re.search(r'\s(\d+)\/(\d+)\/(\d+)', line) 
      counts[ints]['queue'] = int(regx.group(1)) 
      counts[ints]['drops'] = int(regx.group(2)) 
      counts[ints]['no-buffer'] = int(regx.group(3)) 

return counts 

我试图创建一个额外的深度级别的字典，但我对数得到一个关键的错误[整数] [spolicies]线。从我读的内容来看，我认为这是嵌套字典的工作方式，但我想我误解了。

for ints, int_strings in zip(int_names, int_output): 
    counts.setdefault(ints, {}) 

    for line in int_strings.splitlines(): 
     matchpolicy = re.search(r'(Service-policy.*)', line) 
     matchdrops = re.findall(r'total drops.*', line) 
     if matchpolicy: 
      spolicies = matchpolicy.group(0) 
      counts[ints][spolicies] 
     if matchdrops: 
      regx = re.search(r'\s(\d+)\/(\d+)\/(\d+)', line) 
      counts[ints][spolicies]['queue'] = int(regx.group(1)) 
      counts[ints][spolicies]['drops'] = int(regx.group(2)) 
      counts[ints][spolicies]['no-buffer'] = int(regx.group(3)) 

return counts 

无论哪种方式，我猜想有可能是一个更好的方式来组织这个数据，所以我可以以后通过它更容易。有任何想法吗？

Answer 1

labels = ["depth", "drops", "buffer_drops"] 
values = ['0', '14996', '0', '0', '2100', '0'] 
keys=['Gi1','Gi2'] 

values_grouped_by_3 = list(zip(*[iter(values)]*3)) 
data = dict(zip(keys,[dict(zip(labels,vals)) for vals in values_grouped_by_3])) 

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