如何从在for循环中迭代的函数返回值

Question

我正在尝试编写一个程序，该程序将接受用户的输入（在这种情况下，他们从列表中选择同位素），将其与列表进行匹配，然后打印同位素及其半衰期。这一切都很完美，除了我无法使用与 for 循环之外的输入匹配的半衰期。我尝试将它包含在一个函数中以返回结果，但我所做的一切似乎都不起作用。代码如下：

i=0
iso = list(["Oxygen-22", 2.25, "Carbon-15", 2.45, "Flevorium-289", 2.6, "Oxygen-21", 3.42, "Nitrogen-17", 4.17, "Beryllium-14", 4.84, "Nitrogen-16", 7.13, "Oxygen-20", 13.5, "Beryllium-11", 13.8, "Carbon-10", 19.3, "Oxygen-19", 26.5, "Dubnium-261", 27, "Seaborgium-266", 30, "Dubnium-262", 34]);

print("Please select from the below list of isotopes:");
while i < len(iso):
  print iso[i];
  i = i + 2;

def search(x,y):
  u_iso = input("Enter selection:");
  for x in range(len(iso)):
    if u_iso == iso[x]:
      y = x+1;
      print("Isotope:",iso[x],"Half-life:", iso[y],"seconds");
    else:
      x=x+2;
    if u_iso not in iso:
      x=0;
      u_iso = input("Enter selection again:");
      if u_iso == iso[x]:
        y = x+1;
        print("Isotope:",iso[x],"Half-life:", iso[y],"seconds");
      else:
        x=x+2;
  return x, y;

search(0,0);
print x,y;

Answer 1

欢迎来到 Stackoverflow。实际上，您并没有保存函数返回的值，这就是您无法使用它们的原因。 如果你想通过函数获取返回值，你可以将它保存在两个变量中，然后你可以打印出来。

i=0
iso = list(["Oxygen-22", 2.25, "Carbon-15", 2.45, "Flevorium-289", 2.6, "Oxygen-21", 3.42, "Nitrogen-17", 4.17, "Beryllium-14", 4.84, "Nitrogen-16", 7.13, "Oxygen-20", 13.5, "Beryllium-11", 13.8, "Carbon-10", 19.3, "Oxygen-19", 26.5, "Dubnium-261", 27, "Seaborgium-266", 30, "Dubnium-262", 34]);

print("Please select from the below list of isotopes:");
while i < len(iso):
  print iso[i];
  i = i + 2;

def search(x,y):
  u_iso = input("Enter selection:");
  for x in range(len(iso)):
    if u_iso == iso[x]:
      y = x+1;
      print("Isotope:",iso[x],"Half-life:", iso[y],"seconds");
    else:
      x=x+2;
    if u_iso not in iso:
      x=0;
      u_iso = input("Enter selection again:");
      if u_iso == iso[x]:
        y = x+1;
        print("Isotope:",iso[x],"Half-life:", iso[y],"seconds");
      else:
        x=x+2;
  return x, y;

p,q = search(0,0);
print p,q;

如果您有进一步的疑问可以与我分享，我希望它会如您所愿。 ?

Answer 2

你可以试试这个简单的方法....

iso =["Oxygen-22","2.25", "Carbon-15","2.45", "Flevorium-289","2.6", "Oxygen- 
21","3.42", "Nitrogen-17","4.17", "Beryllium-14","4.84", "Nitrogen-16","7.13", 
"Oxygen-20","13.5", "Beryllium-11","13.8", "Carbon-10","19.3", "Oxygen-19", 
"26.5", "Dubnium-261","27", "Seaborgium-266", "30", "Dubnium-262", "34"]
print("Please select from the below list of isotopes: ")
i=0
while i < len(iso):
   print(iso[i])
   i = i + 2
def search():
    u_iso = input("Enter selection: ")
    for x in range(len(iso)):
        if u_iso == iso[x]:
            y = x+1
                print("Isotope:"+iso[x]+"Half-life:"+ iso[y]+"seconds")
            else:
                pass (or) print("Isotope not found")
search()

结果将是..

Enter selection: Oxygen-22
Isotope:Oxygen-22Half-life:2.25seconds
(or)
Enter selection: Iron
Isotope not found

Answer 3

当您有 Key : Value 对时，使用 Dictionary 总是更好 字典上的文档https://www.geeksforgeeks.org/python-dictionary

顺便说一句，我把你的代码转换成了字典，

iso = {}

iso = {"Oxygen-22" : 2.25, "Carbon-15" : 2.45, "Flevorium-289" : 2.6, "Oxygen-21" : 3.42, "Nitrogen-17" : 4.17, "Beryllium-14" : 4.84, "Nitrogen-16" : 7.13, "Oxygen-20" : 13.5, "Beryllium-11" : 13.8, "Carbon-10" : 19.3, "Oxygen-19" : 26.5, "Dubnium-261" : 27, "Seaborgium-266" : 30, "Dubnium-262" : 34}

u_iso = input("Enter selection:");

print(u_iso,end=" ")

print(iso[u_iso])

输出是，

Enter selection: Oxygen-19

Oxygen-19 26.5