查找列表中兩個元素之間關系的最pythonic方法

Question

假設我有以下列表：

l = ["watermelon", "banana", "orange", "apple"]

我想編寫一個函數，該函數返回列表中另一個元素之前的元素。 例如：

>> is_before("banana", "watermelon")
>> False
>> is_before("banana", "apple")
>> True

編寫這樣一個函數的最pythonic 的方法是什么？

Answer 1

你可以這樣做（假設沒有重復）：

l = ["watermelon", "banana", "orange", "apple"]

indeces = {w: i for i, w in enumerate(l)}

def is_previous(x, y):
    return indeces[x] < indeces[y]

>>> is_previous("banana", "watermelon")
False
>>> is_previous("banana", "apple")
True

這不處理任何參數不在初始列表中的情況。

Answer 2

mylist = ["watermelon", "banana", "orange", "apple"]

def is_before(prev_item, target, arr):
    return prev_item in arr[:arr.index(target)]

>>>is_before("banana", "apple", mylist)
True
>>>is_before("banana", "watermelon", mylist)
False

如果你想處理重復，你可以使用這樣的東西

def find_item_last_index(count, item, arr, index=0):
    # A recursive function for finding the last index of an item in a list
    if count == 1:
        return index + arr.index(item)
    return (find_item_last_index(count-1, item, arr[arr.index(item)+1:],  
                                 index+arr.index(item)+1))

def is_before(prev_item, target, arr):
    return prev_item in arr[: find_item_last_index(arr.count(target), target, arr)]

mylist =  ["watermelon", "apple", "banana", "orange", "apple"]

>>>is_before("banana", "apple", mylist)
True
>>>is_before("banana", "watermelon", mylist)
False