두 목록 간의 공통 요소 비교

두 목록 간의 공통 요소 비교

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def common_elements(list1, list2):
    """
    Return a list containing the elements which are in both list1 and list2

    >>> common_elements([1,2,3,4,5,6], [3,5,7,9])
    [3, 5]
    >>> common_elements(['this','this','n','that'],['this','not','that','that'])
    ['this', 'that']
    """
    for element in list1:
        if element in list2:
            return list(element)

지금까지 확인했지만 작동하지 않는 것 같습니다!

어떤 아이디어?

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>>> list1 = [1,2,3,4,5,6]
>>> list2 = [3, 5, 7, 9]
>>> list(set(list1).intersection(list2))
[3, 5]

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집합을 사용하고 한 줄로 공통점을 얻을 수도 있습니다. 집합 중 하나에서 차이점을 포함하는 집합을 뺍니다.

A = [1,2,3,4]
B = [2,4,7,8]
commonalities = set(A) - (set(A) - set(B))

---

제안한 솔루션 S.Mark 및 SilentGhost는 그것이 파이썬 방식으로 수행하는 방법을 일반적으로 말해,하지만 난 당신의 솔루션은 작업을하지 않는 이유를 알고부터 당신은 또한 혜택을 누릴 수있다 생각했다. 문제는 두 목록에서 첫 번째 공통 요소를 찾으면 해당 단일 요소 만 반환한다는 것입니다. result목록 을 만들고 해당 목록에서 공통 요소를 수집하여 솔루션을 수정할 수 있습니다 .

def common_elements(list1, list2):
    result = []
    for element in list1:
        if element in list2:
            result.append(element)
    return result

목록 내포를 사용하는 더 짧은 버전 :

def common_elements(list1, list2):
    return [element for element in list1 if element in list2]

그러나 내가 말했듯이 이것은 이것을 수행하는 매우 비효율적 인 방법입니다. Python의 내장 세트 유형은 내부적으로 C로 구현되기 때문에 훨씬 더 효율적입니다.

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set 교차점 사용, set (list1) & set (list2)

>>> def common_elements(list1, list2):
...     return list(set(list1) & set(list2))
...
>>>
>>> common_elements([1,2,3,4,5,6], [3,5,7,9])
[3, 5]
>>>
>>> common_elements(['this','this','n','that'],['this','not','that','that'])
['this', 'that']
>>>
>>>

결과 목록은 원래 목록과 순서가 다를 수 있습니다.

---

간단한 목록 이해를 사용할 수 있습니다.

x=[1,2,3,4]
y=[3,4,5]
common = [i for i in x if i in y]
common: [3,4]

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이전 답변은 모두 고유 한 공통 요소를 찾기 위해 작동하지만 목록에서 반복되는 항목을 설명하지 못합니다. 공통 요소가 목록에서 공통적으로 발견되는 것과 동일한 번호로 표시되도록하려면 다음 한 줄을 사용할 수 있습니다.

l2, common = l2[:], [ e for e in l1 if e in l2 and (l2.pop(l2.index(e)) or True)]

or True당신이 평가하는 모든 요소를 기대한다면 부분에만 필요합니다 False.

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이것은 for 루프보다 세트가 더 쉽다고 생각하는 내 제안입니다.

def unique_common_items(list1, list2):
   # Produce the set of *unique* common items in two lists.
   return list(set(list1) & set(list2))

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세트는이 문제를 해결할 수있는 또 다른 방법입니다.

a = [3,2,4]
b = [2,3,5]
set(a)&set(b)
{2, 3}

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1) Method1 saving list1 is dictionary and then iterating each elem in list2

def findarrayhash(a,b):
    h1={k:1 for k in a}
    for val in b:
        if val in h1:
            print("common found",val)
            del h1[val]
        else:
            print("different found",val)
    for key in h1.iterkeys():
        print ("different found",key)

Finding Common and Different elements:

2) Method2 using set

def findarrayset(a,b):
    common = set(a)&set(b)
    diff=set(a)^set(b)
    print list(common)
    print list(diff) 

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Use a generator:

common = (x for x in list1 if x in list2)

The advantage here is that this will return in constant time (nearly instant) even when using huge lists or other huge iterables.

For example,

list1 =  list(range(0,10000000))
list2=list(range(1000,20000000))
common = (x for x in list1 if x in list2)

All other answers here will take a very long time with these values for list1 and list2.

You can then iterate the answer with

for i in common: print(i)

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Here's a rather brute force method that I came up with. It's certainly not the most efficient but it's something.

The problem I found with some of the solutions here is that either it doesn't give repeated elements or it doesn't give the correct number of elements when the input order matters.

#finds common elements
def common(list1, list2):
    result = []
    intersect = list(set(list1).intersection(list2))

    #using the intersection, find the min
    count1 = 0
    count2 = 0
    for i in intersect: 
        for j in list1:
            if i == j:
                count1 += 1
        for k in list2: 
            if i == k:
                count2 += 1
        minCount = min(count2,count1)
        count1 = 0
        count2 = 0

        #append common factor that many times
        for j in range(minCount):
            result.append(i)

    return result

---

f_list=[1,2,3,4,5] # First list
s_list=[3,4,5,6,7,8] # Second list
# An empty list stores the common elements present in both the list
common_elements=[]

for i in f_list:
    # checking if each element of first list exists in second list
    if i in s_list:
        #if so add it in common elements list
        common_elements.append(i) 
print(common_elements)

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a_list = range(1,10)
b_list = range(5, 25)
both = []

for i in b_list:
    for j in a_list:
        if i == j:
            both.append(i)

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Hi, this is my propose (very simple)

import random

i = [1,4,10,22,44,6,12] #first random list, could be change in the future
j = [1,4,10,8,15,14] #second random list, could be change in the future
for x in i: 
    if x in j: #for any item 'x' from collection 'i', find the same item in collection of 'j'
        print(x) # print out the results

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def common_member(a, b): 
    a_set = set(a) 
    b_set = set(b) 
    if (a_set & b_set): 
        print(a_set & b_set) 
    else: 
        print("No common elements") 

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list_1=range(0,100)
list_2=range(0,100,5)
final_list=[]
for i in list_1:
    for j in list_2:
        if i==j:
            final_list.append(i)
print(set(final_list))

---

Your problem is that you're returning from inside the for loop so you'll only get the first match. The solution is to move your return outside the loop.

def elementosEnComunEntre(lista1,lista2):

    elementosEnComun = set()

    for e1 in lista1:
         if(e1 in lista2):
             elementosEnComun.add(e1)

    return list(elementosEnComun)

---

def list_common_elements(l_1,l_2,_unique=1,diff=0):
    if not diff:
        if _unique:
            return list(set(l_1)&set(l_2))
        if not _unique:
            return list((i for i in l_1 if i in l_2))
    if diff:
        if _unique:
            return list(set(l_1)^set(l_2))
        if not _unique:
            return list((i for i in l_1 if i not in l_2))
"""
Example:
l_1=             [0, 1, 2, 3, 3, 4, 5]
l_2=             [6, 7, 8, 8, 9, 5, 4, 3, 2]
look for diff
l_2,l_1,diff=1,_unique=1: [0, 1, 6, 7, 8, 9]        sorted(unique(L_2 not in L_1) + unique(L_1 not in L_2))
l_2,l_1,diff=1,_unique=0: [6, 7, 8, 8, 9]           L_2 not in L_1
l_1,l_2,diff=1,_unique=1: [0, 1, 6, 7, 8, 9]        sorted(unique(L_1 not in L_2) + unique(L_2 not in L_1))
l_1,l_2,diff=1,_unique=0: [0, 1]                    L_1 not in L_2
look for same
l_2,l_1,diff=0,_unique=1: [2, 3, 4, 5]              unique(L2 in L1)
l_2,l_1,diff=0,_unique=0: [5, 4, 3, 2]              L2 in L1
l_1,l_2,diff=0,_unique=1: [2, 3, 4, 5]              unique(L1 in L2)
l_1,l_2,diff=0,_unique=0: [2, 3, 3, 4, 5]           L1 in L2
"""

This function provides the ability to compare two lists (L_1 vs. L_2). DIFF parameter set the comparison to search for common elements (DIFF==True) or different elements (DIFF==False) between both lists. On the next level the behavior of the method is set by _UNIQUE parameter. _UNIQUE==True will use Python sets – in this case the method returns a sorted list of unique elements satisfying DIFF. When _UNIQUE==False – the returned list is more explicit, i.e. it will first contain all elements of L_1 followed by all elements of L_2 satisfying the DIFF. Since the output will contain repetitive occurrences of elements in L_1 and L_2 satisfying DIFF, the user could post-count the number of times an element differs or is common between the lists. As this proposal is simply a compilation of code proposed by “cowlinator” and J.S Method2, pleases see these authors posts for discussion on the speed and the performance of the calculation. Credits cowlinator and J.S. Method2

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list1=[1,2,3,4,5]
list2=[4,5,6,3,7]
for c in list1: 
    if(c in list2): 
        print("common elements in list are:",c)
    else:
        print("non common elements in list are:",c)

참고URL : https://stackoverflow.com/questions/2864842/common-elements-comparison-between-2-lists