Python 从一个列表中找出两个加起来就是一个特定数字的数字
这是超级糟糕和混乱,我是新手,请帮助我 基本上,我试着从一个列表中找出两个数字,加起来就是一个目标数字 我用Python 从一个列表中找出两个加起来就是一个特定数字的数字,python,for-loop,permutation,Python,For Loop,Permutation,这是超级糟糕和混乱,我是新手,请帮助我 基本上,我试着从一个列表中找出两个数字,加起来就是一个目标数字 我用lst=[2,4,6,10]和目标值target=8建立了一个示例。本例中的答案是(2,6)和(6,2) 下面是我的代码,但它又长又难看,我相信有更好的方法。你能看看我如何从下面的代码中改进吗 from itertools import product, permutations numbers = [2, 4, 6, 10] target_number = 8 two_nums =
lst=[2,4,6,10]
和目标值target=8
建立了一个示例。本例中的答案是(2,6)
和(6,2)
下面是我的代码,但它又长又难看,我相信有更好的方法。你能看看我如何从下面的代码中改进吗
from itertools import product, permutations
numbers = [2, 4, 6, 10]
target_number = 8
two_nums = (list(permutations(numbers, 2)))
print(two_nums)
result1 = (two_nums[0][0] + two_nums[0][1])
result2 = (two_nums[1][0] + two_nums[1][1])
result3 = (two_nums[2][0] + two_nums[2][1])
result4 = (two_nums[3][0] + two_nums[3][1])
result5 = (two_nums[4][0] + two_nums[4][1])
result6 = (two_nums[5][0] + two_nums[5][1])
result7 = (two_nums[6][0] + two_nums[6][1])
result8 = (two_nums[7][0] + two_nums[7][1])
result9 = (two_nums[8][0] + two_nums[8][1])
result10 = (two_nums[9][0] + two_nums[9][1])
my_list = (result1, result2, result3, result4, result5, result6, result7, result8, result9, result10)
print (my_list)
for i in my_list:
if i == 8:
print ("Here it is:" + str(i))
对于列表中的每一个数字,您都可以查找其补码(当添加到上一个数字时,将给出所需的
目标值
总和)。如果存在,则获取该对并退出,否则继续
如下所示:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers[:-1]): # note 1
complementary = target_number - number
if complementary in numbers[i+1:]: # note 2
print("Solution Found: {} and {}".format(number, complementary))
break
else: # note 3
print("No solutions exist")
done = False
for i, val in enumerate(numbers):
if val >= target_number:
continue
for j, val2 in enumerate(numbers, i+1):
if val + val2 == target_number:
print ("Here it is: " + str(i) + "," + str(j))
done = True
break
if done:
break
产生:
Solution Found: 2 and 6
注:
number[i+1:://code>。之前的数字已经检查过了。切片的一个积极的副作用是,例如列表中存在一个4
,不会为目标值8
提供一对
for
-循环中的else
用法不易理解且经常混淆。只有当循环没有因中断而突然结束时,else
才会触发
如果您可以接受例如
4
-4
解决方案,即使列表中有一个单个4
,您也可以修改如下:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers):
complementary = target_number - number
if complementary in numbers[i:]:
print("Solution Found: {} and {}".format(number, complementary))
break
else:
print("No solutions exist")
对于列表中的每一个数字,您都可以查找其补码(当添加到上一个数字时,将给出所需的
目标值
总和)。如果存在,则获取该对并退出,否则继续
如下所示:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers[:-1]): # note 1
complementary = target_number - number
if complementary in numbers[i+1:]: # note 2
print("Solution Found: {} and {}".format(number, complementary))
break
else: # note 3
print("No solutions exist")
done = False
for i, val in enumerate(numbers):
if val >= target_number:
continue
for j, val2 in enumerate(numbers, i+1):
if val + val2 == target_number:
print ("Here it is: " + str(i) + "," + str(j))
done = True
break
if done:
break
产生:
Solution Found: 2 and 6
注:
number[i+1:://code>。之前的数字已经检查过了。切片的一个积极的副作用是,例如列表中存在一个4
,不会为目标值8
提供一对
for
-循环中的else
用法不易理解且经常混淆。只有当循环没有因中断而突然结束时,else
才会触发
如果您可以接受例如
4
-4
解决方案,即使列表中有一个单个4
,您也可以修改如下:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers):
complementary = target_number - number
if complementary in numbers[i:]:
print("Solution Found: {} and {}".format(number, complementary))
break
else:
print("No solutions exist")
您可以在一行中完成,如下所示:
from itertools import permutations
numbers = [2, 4, 6, 10]
target_number = 8
two_nums = (list(permutations(numbers, 2)))
result=[i for i in two_nums if i[0]+i[1] == target_number]
[(2,6)、(6,2)]
您可以在一行中完成,如下所示:
from itertools import permutations
numbers = [2, 4, 6, 10]
target_number = 8
two_nums = (list(permutations(numbers, 2)))
result=[i for i in two_nums if i[0]+i[1] == target_number]
[(2,6)、(6,2)]
实现这一点的最简单的一般方法是遍历列表,对于每个项目,遍历列表的其余部分,查看它是否与目标值相加。缺点是它是一个O(n^2)操作。我不知道是否有更有效的解决办法。我不能100%确定我的语法是否正确,但它应该如下所示:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers[:-1]): # note 1
complementary = target_number - number
if complementary in numbers[i+1:]: # note 2
print("Solution Found: {} and {}".format(number, complementary))
break
else: # note 3
print("No solutions exist")
done = False
for i, val in enumerate(numbers):
if val >= target_number:
continue
for j, val2 in enumerate(numbers, i+1):
if val + val2 == target_number:
print ("Here it is: " + str(i) + "," + str(j))
done = True
break
if done:
break
当然,您应该将其创建为返回结果的函数,而不仅仅是打印结果。这将消除对“done”变量的需要。实现这一点的最简单的一般方法是迭代列表,并对每个项目迭代列表的其余部分,以查看其总和是否达到目标值。缺点是它是一个O(n^2)操作。我不知道是否有更有效的解决办法。我不能100%确定我的语法是否正确,但它应该如下所示:
numbers = [2, 4, 6, 10]
target_number = 8
for i, number in enumerate(numbers[:-1]): # note 1
complementary = target_number - number
if complementary in numbers[i+1:]: # note 2
print("Solution Found: {} and {}".format(number, complementary))
break
else: # note 3
print("No solutions exist")
done = False
for i, val in enumerate(numbers):
if val >= target_number:
continue
for j, val2 in enumerate(numbers, i+1):
if val + val2 == target_number:
print ("Here it is: " + str(i) + "," + str(j))
done = True
break
if done:
break
当然,您应该将其创建为返回结果的函数,而不仅仅是打印结果。这将消除对“done”变量的需要。列表理解在这里很有效。试试这个:
from itertools import permutations
numbers = [2, 4, 6, 10]
target_number = 8
solutions = [pair for pair in permutations(numbers, 2) if sum(pair) == 8]
print('Solutions:', solutions)
基本上,这个列表理解会查看所有
置换(数字,2)
返回的对,但只保留总和等于8的对。列表理解在这里很有效。试试这个:
from itertools import permutations
numbers = [2, 4, 6, 10]
target_number = 8
solutions = [pair for pair in permutations(numbers, 2) if sum(pair) == 8]
print('Solutions:', solutions)
基本上,此列表理解会查看所有
置换(数字,2)
返回的对,但只保留总和等于8的对。如果您试图找到具有重复值的长列表的多个整数的答案,我建议使用frozenset。“选中”答案将只得到第一个答案,然后停止
import numpy as np
numbers = np.random.randint(0, 100, 1000)
target = 17
def adds_to_target(base_list, target):
return_list = []
for i in range(len(base_list)):
return_list.extend([list((base_list[i], b)) for b in base_list if (base_list[i] + b)==target])
return set(map(frozenset, return_list))
# sample output
{frozenset({7, 10}),
frozenset({4, 13}),
frozenset({8, 9}),
frozenset({5, 12}),
frozenset({2, 15}),
frozenset({3, 14}),
frozenset({0, 17}),
frozenset({1, 16}),
frozenset({6, 11})}
1) 在第一个for循环中,将包含两个与目标值相加的整数的列表添加到“return_list”中,即创建列表列表
2) 然后frozenset取出所有重复对
%timeit adds_to_target(numbers, target_number)
# 312 ms ± 8.86 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
如果您试图找到具有重复值的长列表的多个整数的答案,我建议使用frozenset。“选中”答案将只得到第一个答案,然后停止
import numpy as np
numbers = np.random.randint(0, 100, 1000)
target = 17
def adds_to_target(base_list, target):
return_list = []
for i in range(len(base_list)):
return_list.extend([list((base_list[i], b)) for b in base_list if (base_list[i] + b)==target])
return set(map(frozenset, return_list))
# sample output
{frozenset({7, 10}),
frozenset({4, 13}),
frozenset({8, 9}),
frozenset({5, 12}),
frozenset({2, 15}),
frozenset({3, 14}),
frozenset({0, 17}),
frozenset({1, 16}),
frozenset({6, 11})}
1) 在第一个for循环中,将包含两个与目标值相加的整数的列表添加到“return_list”中,即创建列表列表
2) 然后frozenset取出所有重复对
%timeit adds_to_target(numbers, target_number)
# 312 ms ± 8.86 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
如果您想要一种不用itertools就能高效完成此任务的方法-
numbers = [1,3,4,5,6,2,3,4,1]
target = 5
number_dict = {}
pairs = []
for num in numbers:
number_dict[num] = number_dict.get(num, 0) + 1
complement = target - num
if complement in number_dict.keys():
pairs.append((num, complement))
number_dict.pop(num)
number_dict.pop(complement)
如果您想要一种不用itertools就能高效完成此任务的方法-
numbers = [1,3,4,5,6,2,3,4,1]
target = 5
number_dict = {}
pairs = []
for num in numbers:
number_dict[num] = number_dict.get(num, 0) + 1
complement = target - num
if complement in number_dict.keys():
pairs.append((num, complement))
number_dict.pop(num)
number_dict.pop(complement)
这很简单:)
这很简单:)
在您的示例中,
4
和4
会是一个解决方案吗?这不是更适合或论坛吗?@Ev.Kounis,是的,但在这种情况下,我尽量不按其编号添加。您的代码几乎完全针对问题。如果我们的名单