使用子列表查找列表排列

Question

我正在查看關於遞歸的網站​​ 。當我發現使用遞歸找人名單的排列的這個例子：

def fill_seats(people): 
    # Return a list of ways to fill len(people) seats with the people 
    # named in the sequence people. 
    if len(people) == 0: 
     return [] 
    else: 
     possible = [] 
     for i in range(len(people)): 
     this_person = people[i] 
     everyone_else = people[:i] + people[i+1:] 
     for seating_arrangement in fill_seats(everyone_else): 
      possible = possible + [[this_person] + seating_arrangement] 
     return possible 

our_class = ["Biella", "Gwen", "Katie", "Seth", "Will"] 

for arrangement in fill_seats(our_class): 
    print arrangement 

print len(fill_seats(our_class)), "total possible arrangements." 

但是它使返回0，我不知道爲什麼，任何想法？在這種情況下子串是如何工作的？難道他們不僅僅將單個項目拼接在一起？那有什麼目的？

Answer 1

所有你需要改變的是

if len(people) == 1: 
    return [people] 

因爲，當people是[]，它返回一個空列表。所以，如果人們只有一個元素，fill_seats(everyone_else)將返回[]，所以可能還會返回[]。同樣的東西通過鏈條返回並最終返回。

隨着這種變化，輸出變爲

['Biella', 'Gwen', 'Katie', 'Seth', 'Will'] 
['Biella', 'Gwen', 'Katie', 'Will', 'Seth'] 
['Biella', 'Gwen', 'Seth', 'Katie', 'Will'] 
... 
... 
... 
['Will', 'Seth', 'Gwen', 'Katie', 'Biella'] 
['Will', 'Seth', 'Katie', 'Biella', 'Gwen'] 
['Will', 'Seth', 'Katie', 'Gwen', 'Biella'] 
120 total possible arrangements. 

Answer 2

要回答你的問題有關的切片，這些都不是子，他們的子列表。 例如，如果people是以下列表：

people = ['Adam', 'Betsy', 'Charlie', 'David'] 

我們開始通過在人各項指標迭代。因此，我們的第一次迭代將分配

>>> this_person = people[0] 
>>> this_person 
'Adam' 

然後everyone_else是：

>>> people[:0] 
[] 
>>> people[1:] 
['Betsy', 'Charlie', 'David'] 
>>> everyone_else = people[:0] + people[1:] 
>>> everyone_else 
['Betsy', 'Charlie', 'David'] 

基本上，你重建列表中留出當前指數i，則用較小的列表遞歸。

當i = 1，它看起來像這樣：

>>> this_person = people[1] 
'Betsy' 
>>> people[:1] 
['Adam'] 
>>> people[2:] 
['Charlie', 'David'] 
>>> everyone_else = people[:1] + people[2:] 
>>> everyone_else 
['Adam', 'Charlie', 'David']