如何比較兩個元組？

Question

寫功能commonElements（T1，T2），其在2元組取作爲參數，並返回包含了在兩個元組找到的元素的排序元組。

例子

>>> commonElements((1, 2, 3), (2, 5, 1)) 
(1, 2) 
>>> commonElements((1, 2, 3, 'p', 'n'), (2, 5 ,1, 'p')) 
(1, 2, 'p') 
>>> commonElements((1, 3, 'p', 'n'), ('a', 2 , 5, 1, 'p')) 
(1, 'p') 

我想兩個元

def commonElements(t1,t2): 
    if t1 in t2: 
     return t1 

Answer 1

您將需要檢查，如果每個元素in T1是in T2之間進行比較，你的代碼只是返回t1當你發現第一次共同元件或無如果沒有共同的元素：

def commonElements(t1,t2): 
    temp = [] # store common elements 
    st2 = set(t2) 
    for ele in t1: # loop over each element in t1 
     if ele in st2: # if it is in t2 add it to or temp list, set lookups are O(1) 
      temp.append(ele) 
    return tuple(sorted(temp)) # sort temp and convert to a tuple 

In [4]: commonElements((1, 2, 3), (2, 5, 1)) 
Out[4]: (1, 2) 

In [5]: commonElements((1, 2, 3, 'p', 'n'), (2, 5 ,1, 'p')) 
Out[5]: (1, 2, 'p') 

In [6]: commonElements((1, 3, 'p', 'n'), ('a', 2 , 5, 1, 'p')) 
Out[6]: (1, 'p') 

您還可以使用sets或generator expression：

def commonElements(t1,t2): 
    # find the intersection/common elements 
    return tuple(sorted(set(t1).intersection(t2))) 

def commonElements(t1,t2): 
    return tuple(sorted(ele for ele in t1 if ele in t2)) 

Answer 2

你可以使用的套基於運營商交集。

def commonElements(one, two): 
    return tuple(sorted(set(one) & set(two))) 

commonElements((1,2,3), (2,3,4)) 
# (2, 3)