Question

我试图迭代列表，列表中的变量是字典名称，然后迭代这些字典中的键我的代码是这样的：

IA = {"V":"CVolt","T":"CTemp"}
CLR = {"V":"CLR_Volt","T":"CLR_Temp"}
GT = {"V":"GT_CVolt","T":"GT_Temp"}
PP = ["IA","GT","CLR"]
AFT = {"IA":0,"GT":0,"CLR":0}
AFV = {"IA":0,"GT":0,"CLR":0}
Voltage=0
Vs ={"IA":0.85,"GT":0.85,"CLR":0.85}
Tempreture = 0
Ts ={"IA":30,"GT":30,"CLR":30}
EAK = 7e3
Stress = {"IA":0,"GT":0,"CLR":0}
C = 1
Seff={"IA":0,"GT":0,"CLR":0}
csv.DictReader
with open(File,"r+") as Fin:
    reader = csv.DictReader(Fin, dialect='excel')
    for line in reader:
        for i in PP:
            AFT[i] = math.exp(EAK*((1/Ts[i])-(1/float(line[i["T"]]))))
            AFV[i] = math.exp(float(line[i["T"]])-Vs[i])
            Stress[i] = AFT[i] * AFV[i]
            Seff[i] = Seff[i]+Stress[i]

我的问题是如何获得字典i中Key的值

AFT[i] = math.exp(EAK*((1/Ts[i])-(1/float(line[i["T"]]))))

在这种情况下，在第一个循环中我想获得行[“CTemp”]

欢迎任何建议或不同方法

Answer 1

你可以使用python内置的locals()函数：

>>> IA = {"V":"CVolt","T":"CTemp"}
>>> locals()['IA']
{'T': 'CTemp', 'V': 'CVolt'}
>>> locals()['IA']['T']
'CTemp'

或更好，将你的变量放在一个大字典中：

>>> vars = {
    'IA': {"V":"CVolt","T":"CTemp"},
    'CLR': {"V":"CLR_Volt","T":"CLR_Temp"},
    ...
}
>>> vars['IA']
{'T': 'CTemp', 'V': 'CVolt'}
>>> vars['IA']['T']
'CTemp'

感谢@damienfrancois与当地人完成答案：

math.exp(EAK*((1/Ts[i])-(1/float(line[locals()[i]['T']]))))

与vars dict：

math.exp(EAK*((1/Ts[i])-(1/float(line[vars[i]['T']]))))

Answer 2

您可以像这样更改代码

AFT[i] = math.exp(EAK*((1/Ts[i])-(1/float(line[eval(i)["T"]]))))

请注意使用eval将字符串IA转换为名为IA的对象。

使用locals()或容器词典可能更好;看@ Guy的答案。

从具有字典名称的列表中迭代字典

2 个答案: