一次将一个整数列表输入到程序 1 中，例如：
[1, 3, 1, 4, 4, 3, 1]
任务：
打印一个包含与给定列表完全相同的数字的列表，
但重新排列，以便每个 3 后面紧跟一个 4。3 不能移动索引位置，但每个其他数字都可以移动。
该示例的输出应为：
[1, 3, 4, 1, 1, 3, 4]
到目前为止，我的代码只能完成规则 1 和 2。如何修改我的代码以适应这一点？
newList=[]
n=0
numCount= int(input())
while True:
try:
n = int(input())
except:
break
if len(newList) !=(numCount):
if n == 3:
newList.append(3)
newList.append(4)
else:
newList.append(n)
print(newList)
我建议您首先在输入列表中获取 3 和 4 的所有索引，然后将 3 之后的每个元素与 4 交换。它提供了以下代码，该代码非常简短且易于阅读：
a = [1, 3, 1, 4, 4, 3, 1]
# Get the indexes of 3 and 4 in the list
indexesOf3 = [i for i,elem in enumerate(a) if elem == 3]
indexesOf4 = [i for i,elem in enumerate(a) if elem == 4]
# Swap each element following a 3 with a 4
for i3,i4 in zip(indexesOf3,indexesOf4):
a[i3+1], a[i4] = a[i4], a[i3+1]
print(a)
# [1, 3, 4, 1, 1, 3, 4]
注意：这个代码示例修改了输入列表，但显然它可以很容易地更新为一个返回新列表并保持输入列表不变的函数。
这是一个功能，正是这样做的：
def arrange_list(my_list):
# Copy the whole list
arranged_list=myList.copy()
# Find the all 4s
index_of_4s=[i for i, x in enumerate(arranged_list) if x == 4]
next_4=0
# Loop over the whole list
for i in range(len(arrangedList)):
if(arrangedList[i]==3): # We only care about 3s
# We swap the previously found 4 with a 1
arranged_list[index_of_4s[next_4]]=arranged_list[i+1]
arranged_list[i+1]=4
# We target the next 4
next_4=next_4+1
return arranged_list
如果我们用您的示例对其进行测试，我们会得到：
myList=[1, 3, 1, 4, 4, 3, 1]
arrange_list(myList)
#> [1, 3, 4, 1, 1, 3, 4]
您的问题并没有很好地定义，并且没有考虑到丢失的场景。这段代码的工作很简单，其想法是创建一个新列表。
- 查找输入中 3 的位置
- 在新列表中放置 3，然后是 4
- 放置剩余的元素。
input_list = [1, 3, 1, 4, 4, 3, 1]
# Check the number of 3 and 4
n3 = input_list.count(3)
n4 = input_list.count(4)
if n3 > n4:
for i in range(n3-n4):
input_list.append(4)
elif n4 > n3:
for i in range(n4-n3):
input_list.append(3)
# Now let's look at the order. The 3 should not move and must be followed by a 4.
# Easy way, create a new list.
output_list = [None for i in range(len(input_list))]
# Then I'm using numpy to go faster but the idea is just to extract the ids are which the 3 are placed.
import numpy as np
# Place the 3 and the 4
for elt_3 in np.where(np.asarray(input_list) == 3)[0]:
output_list[elt_3] = 3
output_list[elt_3+1] = 4 # Must be sure that the input_list does not end by a 3 !!!
# Then we don't care of the position for the other numbers.
other_numbers = [x for x in input_list if x != 3 and x != 4]
for i, elt in enumerate(output_list):
if elt is None:
output_list[i] = other_numbers[0]
del other_numbers[0]
print (output_list)
在具有单个循环的更紧凑的版本中，它提供：
input_list = [1, 3, 1, 4, 4, 3, 1]
position_3 = np.where(np.asarray(input_list) == 3)[0]
other_numbers = [x for x in input_list if x != 3 and x != 4] # Not 3 and not 4
output_list = [None for i in range(len(input_list))]
for i, elt in enumerate(output_list):
if elt == 4:
continue
elif i not in position_3:
output_list[i] = other_numbers[0]
del other_numbers[0]
else:
output_list[i] = 3
output_list[i+1] = 4