获得一个错误：列表分配索引超出范围

我正在一个错误列表分配索引超出范围下面的代码：获得一个错误：列表分配索引超出范围

n=input("How many numbers?\n") 
print "Enter ",n," numbers..." 
a=[] 
for i in range(0,n): 
    a[i]=input() 

elem=input("Enter the element to be searched: ") 

for i in range(0,n): 
    if a[i]==elem: 
     flag=1 
     break 

if flag==1: 
    print "Item is present in the list" 
else: 
    print "Item is not present in the list"

来源

2017-04-04 Anik Patra

使用'a.append（）'，'不是A [1] ='。 – zondo

请发布堆栈跟踪。 Python很好，可以告诉你有关错误的详细信息...付出代价！ – tdelaney

添加某种类型的安全性INT，采用列表法追加和运营商在：

n = input("How many numbers?\n") 
n = int(n) 
print "Enter ", n, " numbers..." 
a = [] 
for i in range(n): 
    x = input() 
    a.append(x) 

elem = input("Enter the element to be searched: ") 

if elem in a: 
    print "Item is present in the list" 
else: 
    print "Item is not present in the list"

来源

2017-04-04 16:51:25 brennan

非常感谢。我的问题已解决。 –

你设置列表索引没有它宣称。见：

a=[]

然后，你想要访问一些索引？你正在阅读一个字符串与输入，在使用前转换它。事情会是这样的：

n = int(n) 
a= []*n

来源

2017-04-04 16:52:29 Mikael

使用它像这样，

n=input("How many numbers?\n") 
print "Enter ",n," numbers..." 
# assigning with n times zero values which will get overwritten when you input values. 
a=[0]*n 
for i in range(0,n): 
    a[i]=input() 

elem=input("Enter the element to be searched: ") 

for i in range(0,n): 
    if a[i]==elem: 
     flag=1 
     break 

if flag==1: 
    print "Item is present in the list" 
else: 
    print "Item is not present in the list"

来源

2017-04-04 17:34:36

获得一个错误：列表分配索引超出范围

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