Python 带中断的列表理解

我有下面的代码，我想写在一行与列表理解

list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56]

list3 = []
for i in list1:
    for j in list2:
        if j>i:
            # print(i,j)
            list3.append(j)
            break
print(list1)
print(list3)

输出为：

[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

3.6144596
8.731578200000001

我不知道该把它放在哪里

---

感谢您

首先忽略中断条件有助于构建表达式：

In [32]: [[j for j in list2 if j > i] for i in list1]                                       
Out[32]: 
[[5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [17, 20, 24, 26, 56],
 [24, 26, 56],
 [24, 26, 56],
 [26, 56],
 [56]]

从中可以添加最小约束：

In [33]: [min([j for j in list2 if j > i]) for i in list1]                                  
Out[33]: [5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

---

要构建表达式，首先忽略中断条件有助于：

In [32]: [[j for j in list2 if j > i] for i in list1]                                       
Out[32]: 
[[5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [11, 12, 13, 14, 15, 17, 20, 24, 26, 56],
 [17, 20, 24, 26, 56],
 [24, 26, 56],
 [24, 26, 56],
 [26, 56],
 [56]]

从中可以添加最小约束：

In [33]: [min([j for j in list2 if j > i]) for i in list1]                                  
Out[33]: [5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

---

我试着给AbbeGijly发布的答案计时

结果表明，它比原始解决方案慢。看看吧

import timeit

print(timeit.timeit('''
list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 40, 56]

list3 = []
for i in list1:
    for j in list2:
        if j>i:
            # print(i,j)
            list3.append(j)
            break
'''))

print(timeit.timeit('''
list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 40, 56]
list4 = [[j for j in list2 if j > i] for i in list1]
'''))

输出为：

[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

3.6144596
8.731578200000001

---

我试着给AbbeGijly发布的答案计时

结果表明，它比原始解决方案慢。看看吧

import timeit

print(timeit.timeit('''
list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 40, 56]

list3 = []
for i in list1:
    for j in list2:
        if j>i:
            # print(i,j)
            list3.append(j)
            break
'''))

print(timeit.timeit('''
list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 40, 56]
list4 = [[j for j in list2 if j > i] for i in list1]
'''))

输出为：

[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

3.6144596
8.731578200000001

---

您可以将中断逻辑移动到单独的函数中，然后将该函数放入列表中：

def smallest_value_larger_than_i(candidate_values, i):
    for value in candidate_values:
        if value > i:
            return value
    return None  # Not sure how you want to handle this case

list3 = [smallest_value_larger_than_i(list2, i) for i in list1]

---

这比您原来的解决方案运行得稍慢，但是如果使用列表理解的目标是速度，那么您可以通过改进算法来获得更好的结果。例如，如果两个列表都已排序，则可以在跳过一次后立即丢弃列表2中的元素，而不是对照列表1的其余部分检查它们。您还可以对列表2进行二进制搜索，而不是线性扫描。

您可以将中断逻辑移动到单独的函数中，然后将该函数放入列表中：

def smallest_value_larger_than_i(candidate_values, i):
    for value in candidate_values:
        if value > i:
            return value
    return None  # Not sure how you want to handle this case

list3 = [smallest_value_larger_than_i(list2, i) for i in list1]

---

这比您原来的解决方案运行得稍慢，但是如果使用列表理解的目标是速度，那么您可以通过改进算法来获得更好的结果。例如，如果两个列表都已排序，则可以在跳过一次后立即丢弃列表2中的元素，而不是对照列表1的其余部分检查它们。您还可以对列表2进行二进制搜索，而不是线性扫描。

您无法真正中断列表理解的内部for循环，您可以通过使用next函数查找匹配值的第一个匹配项来避免中断它：

list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56]
list3 = [ next(j for j in list2 if j>i) for i in list1 ]

输出：

print(list1)
print(list3)
[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

如果您担心性能，因为列表理解将比循环慢，您可以使用列表2中的对分搜索来查找下一个更高的值：

from bisect import bisect_left
list3 = [ list2[bisect_left(list2,i+1)] for i in list1 ]

---

这假设list2按升序排序，并且maxlist2>maxlist1

您不能真正中断列表的内部for循环，您可以通过使用next函数查找匹配值的第一个匹配项来避免中断它：

list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56]
list3 = [ next(j for j in list2 if j>i) for i in list1 ]

输出：

print(list1)
print(list3)
[4, 5, 6, 9, 10, 16, 21, 23, 25, 27]
[5, 7, 7, 11, 11, 17, 24, 24, 26, 56]

如果您担心性能，因为列表理解将比循环慢，您可以使用列表2中的对分搜索来查找下一个更高的值：

from bisect import bisect_left
list3 = [ list2[bisect_left(list2,i+1)] for i in list1 ]

这假设list2按升序排序，maxlist2>maxlist1