返回在列表中首次找到元素的索引 - Python

我需要編寫一個函數ind（e，L），它包含一個列表L和一個元素e。函數ind（）應該返回在L中首次找到e的索引。計數從0開始。如果e不是L的元素，則ind（e，L）應該返回等於len（L）的整數。返回在列表中首次找到元素的索引 - Python

這是我到目前爲止有：

def ind(e, L): 
    if e in L: 
     return [L].index('e') 
    if e not in L: 
     return len[L]

有人可以幫我，因爲我不能看着辦吧！

來源

2015-09-19 Benjamin Brooks

做**沒有**編輯您的問題，要求新的o嗯，一旦你吸引了答案。相反，問一個新問題。 – Matt

您需要做一些更改。

刪除大約在L附近的方括號。
刪除e附近的引號，因爲e是一個變量而非值。
添加try,except塊。

代碼：

>>> def ind(e, L): 
     try: 
      return L.index(e) 
     except ValueError: 
      return len(L) 


>>> ind(3, [1,2]) 
2 
>>> ind(3, [1,2,3,4,3]) 
2 
>>> ind('r', ['a']) 
1 
>>> ind('r', ['a', 'r']) 
1 
>>>

來源

2015-09-19 16:53:35

這是低效的......'試試：返回L.index（e）除了ValueError：return len（L）'會更好。 – vaultah

是的，你的代碼有2N的複雜性。編輯N – Pynchia

@vaultah編輯.... –

除了@阿維納什的答案，我建議使用tenary conditional operator是有點簡單：

In [25]: def ind(e, L): 
    ...:  return L.index(e) if e in L else len(L) 

In [26]: lst=[1,2] 

In [27]: ind(2, lst) 
Out[27]: 1 

In [28]: ind(33, lst) 
Out[28]: 2

或者試試評論什麼@vaultah：

In [43]: def ind2(e, L): 
    ...:  try: 
    ...:   return L.index(e) 
    ...:  except ValueError: 
    ...:   return len(L) 
    ...:

以基準測試：

In [65]: s='Python is a dynamic and strongly typed programming language that is designed to emphasize usability. Two similar but incompatible versions of Python are in widespread use (2 and 3). Please consider using [python-2.7] or [python-3.x] tags for version-specific questions about Python.' 

In [66]: lst=list(s) 

In [67]: %timeit ind('r', lst) 
The slowest run took 6.81 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 989 ns per loop 

In [68]: %timeit ind2('r', lst) 
The slowest run took 5.01 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 640 ns per loop 

In [69]: lst2=list(s.replace('r', '')) #remove all 'r's in the list 

In [70]: %timeit ind('r', lst2) 
100000 loops, best of 3: 3.77 µs per loop 

In [71]: %timeit ind2('r', lst2) 
The slowest run took 4.12 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 5.61 µs per loop 

In [72]:

注意該try-except邏輯並不總是更有效

來源

2015-09-19 16:58:58 zhangxaochen

或者說，不引入異常處理或呼喚Python的list.index方法：

def ind(e, L): 
    for index, item in enumerate(L): 
     if item == e: 
      return index 

    return index+1

來源

2015-09-19 17:32:14 TessellatingHeckler

此代碼應工作：

def ind(e, L): 
    if e in L: 
     return L.index(e) 
    if e not in L: 
     return len (L)

來源

2016-08-26 11:53:04

返回在列表中首次找到元素的索引 - Python

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