如果存在列表索引,请执行X

问题:如果存在列表索引,请执行X

在我的程序中,用户输入number n,然后输入n字符串数,这些字符串存储在列表中。

我需要进行编码,以便如果存在某个列表索引,然后运行一个函数。

我已经嵌套了if语句,这使情况变得更加复杂len(my_list)

这是我现在所拥有的简化版本,无法使用:

n = input ("Define number of actors: ")

count = 0

nams = []

while count < n:
    count = count + 1
    print "Define name for actor ", count, ":"
    name = raw_input ()
    nams.append(name)

if nams[2]: #I am trying to say 'if nams[2] exists, do something depending on len(nams)
    if len(nams) > 3:
        do_something
    if len(nams) > 4
        do_something_else

if nams[3]: #etc.

In my program, user inputs number n, and then inputs n number of strings, which get stored in a list.

I need to code such that if a certain list index exists, then run a function.

This is made more complicated by the fact that I have nested if statements about len(my_list).

Here’s a simplified version of what I have now, which isn’t working:

n = input ("Define number of actors: ")

count = 0

nams = []

while count < n:
    count = count + 1
    print "Define name for actor ", count, ":"
    name = raw_input ()
    nams.append(name)

if nams[2]: #I am trying to say 'if nams[2] exists, do something depending on len(nams)
    if len(nams) > 3:
        do_something
    if len(nams) > 4
        do_something_else

if nams[3]: #etc.

回答 0

使用列表的长度len(n)来告知您的决定而不是检查n[i]每个可能的长度,对您来说更有用吗?

Could it be more useful for you to use the length of the list len(n) to inform your decision rather than checking n[i] for each possible length?


回答 1

我需要进行编码,以便如果存在某个列表索引,然后运行一个函数。

这是try块的完美用法:

ar=[1,2,3]

try:
    t=ar[5]
except IndexError:
    print('sorry, no 5')   

# Note: this only is a valid test in this context 
# with absolute (ie, positive) index
# a relative index is only showing you that a value can be returned
# from that relative index from the end of the list...

但是,根据定义,Python列表中0和之间的所有项都len(the_list)-1存在(即,无需尝试,除非您知道0 <= index < len(the_list))。

如果希望索引在0和最后一个元素之间,可以使用enumerate

names=['barney','fred','dino']

for i, name in enumerate(names):
    print(i + ' ' + name)
    if i in (3,4):
        # do your thing with the index 'i' or value 'name' for each item...

如果您正在寻找一些明确的“索引”思想,那么我想您是在问一个错误的问题。也许您应该考虑使用映射容器(例如dict)与序列容器(例如列表)。您可以这样重写代码:

def do_something(name):
    print('some thing 1 done with ' + name)

def do_something_else(name):
    print('something 2 done with ' + name)        

def default(name):
    print('nothing done with ' + name)     

something_to_do={  
    3: do_something,        
    4: do_something_else
    }        

n = input ("Define number of actors: ")
count = 0
names = []

for count in range(n):
    print("Define name for actor {}:".format(count+1))
    name = raw_input ()
    names.append(name)

for name in names:
    try:
        something_to_do[len(name)](name)
    except KeyError:
        default(name)

像这样运行:

Define number of actors: 3
Define name for actor 1: bob
Define name for actor 2: tony
Define name for actor 3: alice
some thing 1 done with bob
something 2 done with tony
nothing done with alice

您也可以使用.get方法而不是try / except来获得较短的版本:

>>> something_to_do.get(3, default)('bob')
some thing 1 done with bob
>>> something_to_do.get(22, default)('alice')
nothing done with alice

I need to code such that if a certain list index exists, then run a function.

This is the perfect use for a try block:

ar=[1,2,3]

try:
    t=ar[5]
except IndexError:
    print('sorry, no 5')   

# Note: this only is a valid test in this context 
# with absolute (ie, positive) index
# a relative index is only showing you that a value can be returned
# from that relative index from the end of the list...

However, by definition, all items in a Python list between 0 and len(the_list)-1 exist (i.e., there is no need for a try, except if you know 0 <= index < len(the_list)).

You can use enumerate if you want the indexes between 0 and the last element:

names=['barney','fred','dino']

for i, name in enumerate(names):
    print(i + ' ' + name)
    if i in (3,4):
        # do your thing with the index 'i' or value 'name' for each item...

If you are looking for some defined ‘index’ thought, I think you are asking the wrong question. Perhaps you should consider using a mapping container (such as a dict) versus a sequence container (such as a list). You could rewrite your code like this:

def do_something(name):
    print('some thing 1 done with ' + name)

def do_something_else(name):
    print('something 2 done with ' + name)        

def default(name):
    print('nothing done with ' + name)     

something_to_do={  
    3: do_something,        
    4: do_something_else
    }        

n = input ("Define number of actors: ")
count = 0
names = []

for count in range(n):
    print("Define name for actor {}:".format(count+1))
    name = raw_input ()
    names.append(name)

for name in names:
    try:
        something_to_do[len(name)](name)
    except KeyError:
        default(name)

Runs like this:

Define number of actors: 3
Define name for actor 1: bob
Define name for actor 2: tony
Define name for actor 3: alice
some thing 1 done with bob
something 2 done with tony
nothing done with alice

You can also use .get method rather than try/except for a shorter version:

>>> something_to_do.get(3, default)('bob')
some thing 1 done with bob
>>> something_to_do.get(22, default)('alice')
nothing done with alice

回答 2

len(nams)应该等于n您的代码。所有索引都0 <= i < n“存在”。

len(nams) should be equal to n in your code. All indexes 0 <= i < n “exist”.


回答 3

只需使用以下代码即可完成:

if index < len(my_list):
    print(index, 'exists in the list')
else:
    print(index, "doesn't exist in the list")

It can be done simply using the following code:

if index < len(my_list):
    print(index, 'exists in the list')
else:
    print(index, "doesn't exist in the list")

回答 4

我需要进行编码,以便如果存在某个列表索引,然后运行一个函数。

您已经知道如何对此进行测试,并且实际上已经在代码中执行了这种测试

对于长的列表,有效索引n0通过n-1包容性。

因此,i 当且仅当列表的长度至少为时,列表才具有索引i + 1

I need to code such that if a certain list index exists, then run a function.

You already know how to test for this and in fact are already performing such tests in your code.

The valid indices for a list of length n are 0 through n-1 inclusive.

Thus, a list has an index i if and only if the length of the list is at least i + 1.


回答 5

使用列表的长度是检查索引是否存在的最快解决方案:

def index_exists(ls, i):
    return (0 <= i < len(ls)) or (-len(ls) <= i < 0)

这还将测试负索引以及具有长度的大多数序列类型(如rangesstr)。

如果无论如何您以后都需要访问该索引处的项目,则宽恕要比权限容易,并且它也更快,更Python化。使用try: except:

try:
    item = ls[i]
    # Do something with item
except IndexError:
    # Do something without the item

这与以下情况相反:

if index_exists(ls, i):
    item = ls[i]
    # Do something with item
else:
    # Do something without the item

Using the length of the list would be the fastest solution to check if an index exists:

def index_exists(ls, i):
    return (0 <= i < len(ls)) or (-len(ls) <= i < 0)

This also tests for negative indices, and most sequence types (Like ranges and strs) that have a length.

If you need to access the item at that index afterwards anyways, it is easier to ask forgiveness than permission, and it is also faster and more Pythonic. Use try: except:.

try:
    item = ls[i]
    # Do something with item
except IndexError:
    # Do something without the item

This would be as opposed to:

if index_exists(ls, i):
    item = ls[i]
    # Do something with item
else:
    # Do something without the item

回答 6

如果要迭代插入的actor数据:

for i in range(n):
    if len(nams[i]) > 3:
        do_something
    if len(nams[i]) > 4:
        do_something_else

If you want to iterate the inserted actors data:

for i in range(n):
    if len(nams[i]) > 3:
        do_something
    if len(nams[i]) > 4:
        do_something_else

回答 7

好的,所以我认为这实际上是可能的(出于参数的目的):

>>> your_list = [5,6,7]
>>> 2 in zip(*enumerate(your_list))[0]
True
>>> 3 in zip(*enumerate(your_list))[0]
False

ok, so I think it’s actually possible (for the sake of argument):

>>> your_list = [5,6,7]
>>> 2 in zip(*enumerate(your_list))[0]
True
>>> 3 in zip(*enumerate(your_list))[0]
False

回答 8

您可以尝试这样的事情

list = ["a", "b", "C", "d", "e", "f", "r"]

for i in range(0, len(list), 2):
    print list[i]
    if len(list) % 2 == 1 and  i == len(list)-1:
        break
    print list[i+1];

You can try something like this

list = ["a", "b", "C", "d", "e", "f", "r"]

for i in range(0, len(list), 2):
    print list[i]
    if len(list) % 2 == 1 and  i == len(list)-1:
        break
    print list[i+1];

回答 9

Oneliner:

do_X() if len(your_list) > your_index else do_something_else()  

完整示例:

In [10]: def do_X(): 
    ...:     print(1) 
    ...:                                                                                                                                                                                                                                      

In [11]: def do_something_else(): 
    ...:     print(2) 
    ...:                                                                                                                                                                                                                                      

In [12]: your_index = 2                                                                                                                                                                                                                       

In [13]: your_list = [1,2,3]                                                                                                                                                                                                                  

In [14]: do_X() if len(your_list) > your_index else do_something_else()                                                                                                                                                                      
1

仅用于信息。恕我直言,try ... except IndexError是更好的解决方案。

Oneliner:

do_X() if len(your_list) > your_index else do_something_else()  

Full example:

In [10]: def do_X(): 
    ...:     print(1) 
    ...:                                                                                                                                                                                                                                      

In [11]: def do_something_else(): 
    ...:     print(2) 
    ...:                                                                                                                                                                                                                                      

In [12]: your_index = 2                                                                                                                                                                                                                       

In [13]: your_list = [1,2,3]                                                                                                                                                                                                                  

In [14]: do_X() if len(your_list) > your_index else do_something_else()                                                                                                                                                                      
1

Just for info. Imho, try ... except IndexError is better solution.


回答 10

不要在方括号前留任何空间。

例:

n = input ()
         ^

提示:您应该在代码上方和/或下方添加注释。不隐藏您的代码。


祝你今天愉快。

Do not let any space in front of your brackets.

Example:

n = input ()
         ^

Tip: You should add comments over and/or under your code. Not behind your code.


Have a nice day.


回答 11

很多答案,而不是简单的答案。

要检查字典dict是否存在索引“ id”:

dic = {}
dic['name'] = "joao"
dic['age']  = "39"

if 'age' in dic

如果存在“年龄”,则返回true。

A lot of answers, not the simple one.

To check if a index ‘id’ exists at dictionary dict:

dic = {}
dic['name'] = "joao"
dic['age']  = "39"

if 'age' in dic

returns true if ‘age’ exists.