在 Python 中比较两个字典之间的整数值

Question

此脚本的目标是将客户端软件的当前版本与供应商提供的当前软件版本进行比较。在这个阶段，我只想打印出可用的内容。

我有两个字典设置：

versions = {
    'v80':[80.24,80.16,80.15,80.7],
    'v81':[81.12,81.7,81.4,81.11],
    'v82':[82.7,82.5,82.9,82.6],
    'v83':[83.0,83.1,83.1,83.0]
}
     
client = {
    'client_1':[80.1,80.1,80.1,80.1],
    'client_2':[81.1,80.1,80.1,80.1],
    'client_3':[82.1,80.1,80.1,80.1],
    'client_4':[81.1," ",80.1,80.1],
    'client_5':[80.1," "," "," "]
}

版本 = 可用/客户端 = 客户端正在运行的版本

有 4 个键值代表 4 个不同的应用程序及其所在的版本。

我尝试了几种方法，但都碰壁了。

我想将client_[n] 中的键值与相应版本进行比较（不确定如何将client_3 与v83、client1 与v80 等联系起来）。比较只是如下：

if client['client_1'][0] < versions['v80'][0]
    print(client.client_1[0],"-->",versions.v80[0])

Answer 1

您可以使用以下 for 循环：

>>> for i in range(len(versions)):
...     for j in range(len(versions['v8'+str(i)])):
...         try:
...             if client['client_' + str(i+1)][j] < versions['v8' + str(i)][j]:
...                 print ("client_" + str(i+1), "Program_" + str(j+1), client["client_" + str(i+1)][j],"-->",versions["v8"+str(i)][j])
...         except TypeError:
...             print ("client_" + str(i+1), "Program_" + str(j+1), "Missing")
...             continue
... 
client_1 Program_1 80.1 --> 80.24
client_1 Program_2 80.1 --> 80.16
client_1 Program_3 80.1 --> 80.15
client_1 Program_4 80.1 --> 80.7
client_2 Program_1 81.1 --> 81.12
client_2 Program_2 80.1 --> 81.7
client_2 Program_3 80.1 --> 81.4
client_2 Program_4 80.1 --> 81.11
client_3 Program_1 82.1 --> 82.7
client_3 Program_2 80.1 --> 82.5
client_3 Program_3 80.1 --> 82.9
client_3 Program_4 80.1 --> 82.6
client_4 Program_1 81.1 --> 83.0
client_4 Program_2 Missing
client_4 Program_3 80.1 --> 83.1
client_4 Program_4 80.1 --> 83.0

但是，这取决于您提供的信息是否一致，versions 和clients 的数量相同，列表长度相同。

如果你更清楚一点，我可以尝试一个更强大的解决方案。

Answer 2

您可以引入作为接口的第三个字典

versions = {
'v80':[80.24,80.16,80.15,80.7],
'v81':[81.12,81.7,81.4,81.11],
'v82':[82.7,82.5,82.9,82.6],
'v83':[83.0,83.1,83.1,83.0]
}

interface = {

"client_1":"v80",
"client_2":"v81",
"client_3":"v82",
"client_4":"v83",
"client_5":"v84"
}

client = {
'client_1':[80.1,80.1,80.1,80.1],
'client_2':[81.1,80.1,80.1,80.1],
'client_3':[82.1,80.1,80.1,80.1],
'client_4':[81.1," ",80.1,80.1],
'client_5':[80.1," "," "," "]
}

现在只需使用以下内容进行比较：

if client['client_1'][0] < versions[interface["client_1"]][0]:
    print client["client_1"][0],"-->",versions[interface["client_1"]][0]

Answer 3

你可以试试：

if client['client_1'][0] < versions['v80'][0]:
    print(client["client_1"][0] ,"-->",versions["v80"][0])

为所有客户端和版本执行此操作：

versions = {
'v80':[80.24,80.16,80.15,80.7],
'v81':[81.12,81.7,81.4,81.11],
'v82':[82.7,82.5,82.9,82.6],
'v83':[83.0,83.1,83.1,83.0]
}

client = {
'client_1':[80.1,80.1,80.1,80.1],
'client_2':[81.1,80.1,80.1,80.1],
'client_3':[82.1,80.1,80.1,80.1],
'client_4':[81.1," ",80.1,80.1],
'client_5':[80.1," "," "," "]
}


for cli, num_cl in client.items():
    for ver, num in versions.items():
        if int(num[0]) == int(num_cl[0]):
            print(cli, num_cl, ver, num)

还有输出：

client_4 [81.1, ' ', 80.1, 80.1] v81 [81.12, 81.7, 81.4, 81.11]
client_5 [80.1, ' ', ' ', ' '] v80 [80.24, 80.16, 80.15, 80.7]
client_2 [81.1, 80.1, 80.1, 80.1] v81 [81.12, 81.7, 81.4, 81.11]
client_3 [82.1, 80.1, 80.1, 80.1] v82 [82.7, 82.5, 82.9, 82.6]
client_1 [80.1, 80.1, 80.1, 80.1] v80 [80.24, 80.16, 80.15, 80.7]