Python将更改的元素从一个列表返回到另一个列表的最快方法

Question

我有以下代码：

list_a = [1,2,3,4,5,6,7,8]
list_b = [1,2,3,4,999,6,7,8]

for i in range(len(list_b)):
    if (list_a[i] != list_b[i]):
        print('The element in the original list has changed to: ' + str(list_b[i]))

我的问题：有没有更有效的方法来做到这一点？ 两个列表中数字的顺序无关紧要（但在这种情况下，它取决于此）； 我只想在list_b的元素在list_a中发生更改时能够返回

编辑：确保的因素：

• 列表总是相同的长度
• List_b只会改变一个元素

一些功能：

return_change_in_list(list_a, list_b)
>>> '999' 

Answer 1

使用列表理解的方法（这不是最好的，我接下来将解释）：

list_a = [1,2,3,4,5,6,7,8]
list_b = [1,2,3,4,999,6,7,8]

result = [list_b[i] for i in range(len(list_b)) if list_a[i]!=list_b[i]]
if result:
  print("The element in the original list has changed to: %d." % result[0])

输出：

The element in the original list has changed to: 999.

让我们使用timeit测量五个不同函数的运行时间，以找出最快的方法：

import timeit

def return_change_in_list1(list_a, list_b):
  """function 1, similar to the presented in the question"""
  for i in range(len(list_b)):
    if (list_a[i] != list_b[i]):
      return list_b[i]

def return_change_in_list2(list_a, list_b):
  """function 2"""
  if list_a==list_b:
    return
  return [list_b[i] for i in range(len(list_b)) if list_a[i]!=list_b[i]][0]

def return_change_in_list3(list_a, list_b):
  """function 3"""
  result = [list_b[i] for i in range(len(list_b)) if list_a[i]!=list_b[i]]
  if result:
    return result[0]

def return_change_in_list4(list_a, list_b):
  """function 4, answer of Jab"""
  for a, b in zip(list_a, list_b):
    if a != b:
      return b

def return_change_in_list5(list_a, list_b):
  """function 5, answer of GZ0 and Jab"""
  return next((b for a, b in zip(list_a, list_b) if a != b), None)

list1 = [i for i in range(10000)]+[10001]+[i for i in range(10000)]  # I could also put list1=[i for i in range(10000)]+[10001]+[i for i in range(10001,20001)]...
list2 = [i for i in range(10000)]+[1]+[i for i in range(10000)]

print("function 1;          function 2;          function 3;")
for i in range(10):
  print(timeit.timeit('return_change_in_list1(list1,list2)', globals=globals(), number=1000), end='; ')
  print(timeit.timeit('return_change_in_list2(list1,list2)', globals=globals(), number=1000), end='; ')
  print(timeit.timeit('return_change_in_list3(list1,list2)', globals=globals(), number=1000), end=';\n')
print("\n\nfunction 4;          function 5")
for i in range(10):  
  print(timeit.timeit('return_change_in_list4(list1,list2)', globals=globals(), number=1000), end='; ')
  print(timeit.timeit('return_change_in_list5(list1,list2)', globals=globals(), number=1000))

在我的电脑中的输出之一：

function 1;          function 2;          function 3;
4.1352741499999865; 7.948646284999995; 6.719404378000036;
3.991087896999943; 7.604543525999816; 7.098497474999931;
4.010241690999919; 7.593738104000295; 7.702650748999986;
4.19843784700015; 7.672624406000068; 7.5358697430001484;
4.14458595699989; 9.280372852000255; 8.256589966999854;
4.391348879000361; 7.764564555000106; 7.068202192999706;
4.2762628720001885; 8.209926066000207; 6.952296369000123;
3.8595704109998223; 11.935936167999898; 16.16772019300015;
8.138206416999765; 12.870063486000163; 15.746456568000212;
7.82583780899995; 11.736876751000182; 8.441530435000004;


function 4;          function 5
3.098060323000027; 2.7393256840000504
2.7861139729998285; 2.548324536000109
2.5771596439999485; 2.5661135760001343
2.7238574630000585; 2.594752339000024
2.6554647250000016; 2.8191915849997713
2.6247645640000883; 2.84076504199993
3.3453383609999037; 2.874788166000144
2.934053877000224; 2.950007811999967
5.239812221999728; 5.56969765399981
6.443700938999882; 4.463028083000154

当我将列表的值更改为：

list1=[i for i in range(20000)]+[10001]
list2=[i for i in range(20000)]+[1]

输出是：

function 1;          function 2;          function 3;
7.649340118000055; 7.448747982999976; 6.952490765999755;
7.988774744000239; 7.504327044999627; 6.845772889999807;
8.663970897000127; 7.184423781000078; 6.306836058000044;
8.560694254999817; 9.84430615500014; 7.21160759400027;
9.264006450000124; 10.213202504000037; 6.9900130920000265;
8.118536243000108; 7.994644583000081; 8.72104606899984;
9.391442627000288; 10.152986846000204; 9.21849292800016;
12.79642554100019; 18.673425052999846; 9.258570297000006;
9.084169438000117; 7.639437806999922; 6.252692509000099;
8.07683060699992; 8.637334575000295; 6.861809983000057;


function 4;          function 5
5.401317727999867; 5.694436690000202
5.218023895999977; 5.176447093999741
5.217633329999899; 6.5157359929999075
5.668167091000214; 5.585341486999823
5.535178367999833; 5.495608334000281
5.182497067999975; 5.231392088000121
5.375581145000069; 5.422244396000224
6.115360925999994; 6.473182661999999
6.920992358000149; 5.24543598199989
5.399142923999989; 5.373359425000217

当我将列表的值更改为：

list1=[10001]+[i for i in range(20000)]
list2=[1]+[i for i in range(20000)]

输出是：

function 1;          function 2;          function 3;
0.0009529700000712182; 6.419712152000102; 6.169585547999759;
0.0013693759997295274; 6.107828072999837; 7.524482284999976;
0.0012241990002621606; 6.5767201450003085; 6.710827992999839;
0.0014647069997408835; 7.1059040999998615; 7.801625528999921;
0.002128201000232366; 9.78993420400002; 11.612126786999852;
0.0013582339997810777; 8.097604090000004; 7.4285691270001735;
0.0017677549999461917; 7.782655680000062; 6.897861279999688;
0.0014854589999231393; 6.621780061000209; 6.887953365999692;
0.0013960719998067361; 6.920186684999862; 6.5975830910001605;
0.0014639700002589962; 7.075127180999971; 6.68892119200018;


function 4;          function 5
0.000932528000248567; 0.0016549910001231183
0.0010366090000388795; 0.0016346869997505564
0.000914027999897371; 0.0015974439997989975
0.001006645999950706; 0.0018295350000698818
0.0010002259996326757; 0.001662419999775011
0.0009906569998747727; 0.0016710579998289177
0.0011088470000686357; 0.0016620189999230206
0.0010634639997988415; 0.0018417569999655825
0.0008912280000004102; 0.001735908000227937
0.0009644479996495647; 0.0015510050002376374

注意，不同的列表会导致不同的运行时间，不同的range也可能会影响结果，请尝试自己计时！

似乎函数 4 和 5 是这五个中最快的。

列表list_a需要遍历整个list_a和list_b 。 当只需要第一个元素时，会有很多浪费的计算。” - GZ0

确实如此，函数 4 和 5 似乎是要走的路。 :)

请注意，功能 4 基于Jab的回答，而功能 5 基于GZ0的回答和Jab的回答。 这个答案花了我很多时间，但我认为这是值得的。

Answer 2

您可以检查每次迭代的每个元素是否相同另外itertools.zip_longest将确保迭代，直到至少最小的列表用完

list_a = [1,2,3,4,5,6,7,8]
list_b = [1,2,3,4,999,6,7,8]

def find_diff(list_a, list_b):
  for a, b in zip(list_a, list_b):
    if a != b:
      return b
  return None


print(find_diff(list_a, list_b))

这将返回999 。 如果所有元素都相同，则返回None

这也可以像 GZO 所说的那样使用next来完成，因为它与我的逻辑相同。 但我会使用None作为默认值。

next((b for a, b in zip(list_a, list_b) if a != b), None)

Answer 3

next(b for a, b in zip(list_a, list_b) if a != b)