[英]What's the most efficient way to perform a multiple match lookup in a python dictionary?
[英]What is the most efficient way to a multiple variable in dictionary in python?
我正在寻找的这是我的代码,这是以其他最有效的方式进行编码的方法吗? 我有多个变量,并插入到字典中。 请提出建议,其他选项(例如array等)也可以。
def momentEndSpan(span_type,max_combo,length):
if "simply supported" == span_type:
q = max_combo
force = {}
RA = {"PA" : q*length/2}
RB = {"PB" : q*length/2}
RA_moment = {"MA" : 0}
R_mid_moment = {"Mmid": (q*math.pow(length,2))/8 }
RB_moment = { "MB" : 0}
force.update(RA)
force.update(RB)
force.update(RA_moment)
force.update(R_mid_moment)
force.update(RB_moment)
return force
elif "one end continuous" == span_type:
q = max_combo
x = (3/8)*length
force = {}
RA = {"Phinge" : 3*q*length/8}
RB = {"Pfixed" : 5*q*length/8}
RA_moment = {"Mhinge" : 0}
R_mid_moment = {"Mmid": (q*math.pow(length,2))*(9/128) }
RB_moment = { "MB" : -1*(q*math.pow(length,2))/8 }
force.update(RA)
force.update(RB)
force.update(RA_moment)
force.update(R_mid_moment)
force.update(RB_moment)
return force
非常感谢你
“更多Python语言”方法是创建一个字典并更新一次。
q = max_combo
force = {}
if "simply supported" == span_type:
new = {"PA" : q*length/2,
"PB" : q*length/2,
"MA" : 0, "Mmid": (q*math.pow(length,2))/8,
"MB" : 0}
elif "one end continuous" == span_type:
x = (3/8)*length
new = {"Phinge" : 3*q*length/8,
"Pfixed" : 5*q*length/8,
"Mhinge" : 0,
"Mmid": (q*math.pow(length,2))*(9/128),
"MB" : -1*(q*math.pow(length,2))/8 }
force.update(new)
另外,请注意,如果force
字典不包含任何先前定义的项目,则可以简单地返回new
和/或在有任何后续操作的情况下继续更新new
。 或者只使用name force
而不是new
。
q = max_combo
if "simply supported" == span_type:
force = {...}
elif "one end continuous" == span_type:
x = (3/8)*length
force = {...}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.