使用 merge sort/quick sort 对 Python 中 class 个对象的属性进行排序
Using merge sort/quick sort to sort attribute of class objects in Python
我有这个学生class
class Student:
def __init__(self, name, id):
self.name = name
self.id = id
我需要对 Student class 的一些对象进行排序,特别是使用 merge/quick 排序 对 id
和预期结果进行排序是学生姓名的数组。所以如果我有这些对象:
s1 = Student("Andy", 4)
s2 = Student("Bob", 3)
s3 = Student("Sophie", 2)
s4 = Student("Tony", 1)
s5 = Student("Jerry", 5)
预期结果:
result = ["Tony", "Sophie", "Bob", "Andy", "Jerry"]
我不确定是否需要创建对象数组或放置排序函数的位置。
有什么想法吗?
将其转换成对数组,然后应用合并排序并将数组作为参数传递
def merge(arr, l, m, r):
n1 = m - l + 1
n2 = r- m
# create temp arrays
L = [0] * (n1)
R = [0] * (n2)
# Copy data to temp arrays L[] and R[]
for i in range(0 , n1):
L[i] = arr[l + i]
for j in range(0 , n2):
R[j] = arr[m + 1 + j]
# Merge the temp arrays back into arr[l..r]
i = 0 # Initial index of first subarray
j = 0 # Initial index of second subarray
k = l # Initial index of merged subarray
enter code here
while i < n1 and j < n2 :
if L[i][1] <= R[j][1]: // here you need to compare id
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Copy the remaining elements of L[], if there
# are any
while i < n1:
arr[k] = L[i]
i += 1
k += 1
# Copy the remaining elements of R[], if there
# are any
while j < n2:
arr[k] = R[j]
j += 1
k += 1
def mergeSort(arr,l,r):
if l < r:
# Same as (l+r)//2, but avoids overflow for
# large l and h
m = (l+(r-1))//2
# Sort first and second halves
mergeSort(arr, l, m)
mergeSort(arr, m+1, r)
merge(arr, l, m, r)
arr = [( "sdas",1), ( "asd3",3), ( "asd1",2)]
n = len(arr)
print ("Given array is")
for i in range(n):
print ("%s" %arr[i][0]),
mergeSort(arr,0,n-1)
print ("\n\nSorted array is")
for i in range(n):
print (arr[i]),
#or you can use predefined sort function like this
def sortSecond(val):
return val[1]
arr = [( "sdas",1), ( "asd3",3), ( "asd1",2)]
# sorts the array in ascending according to
# second element
arr.sort(key = sortSecond)
print(arr)
您可以使用 is 创建学生列表。
假设你将自己编写归并排序函数,现在你有一个这样的函数定义:
def merge_sort(students_array)
现在应该以特定方式实现此功能,您可以根据 id 比较元素。
为此,您有几种选择:
- 有一个__eq__(函数支持a == b),__le__(函数支持a <= b),__ge__(函数支持a >= b ), __gt__ (函数支持a > b), __lt__ (函数支持a < b)。您可以只使用“<、>、=”IMO,但这些是您要考虑的功能。 Student 需要实现它们,然后可以通过比较他们的 ID 来比较两个学生 - 这可能是最 pythonic 的方法:
示例实现:
class Student:
def __init__(self, name, id):
self.name = name
self.id = id
def __eq__(self, other):
return self.id == other.id
def __gt__(self, other):
return self.id > other.id
def __lt__(self, other):
return self.id < other.id
...
# getting a list of Students, not only ids
def merge_sort(students_array):
...
# example compare in the sort function
# this will be translated to: students_array[i].__gt__(students_array[j])
# which will return: students_array[i] > students_array[j]
if students_array[i] > students_array[j]:
# do something
...
def main():
s1 = Student("Andy", 4)
s2 = Student("Bob", 3)
s3 = Student("Sophie", 2)
s4 = Student("Tony", 1)
s5 = Student("Jerry", 5)
students_array = [s1, s2, s3, s4, s5]
merge_sort(students_array)
关于 python dunder 函数的良好文档:
https://docs.python.org/3/library/operator.html
希望有所帮助
我有这个学生class
class Student:
def __init__(self, name, id):
self.name = name
self.id = id
我需要对 Student class 的一些对象进行排序,特别是使用 merge/quick 排序 对 id
和预期结果进行排序是学生姓名的数组。所以如果我有这些对象:
s1 = Student("Andy", 4)
s2 = Student("Bob", 3)
s3 = Student("Sophie", 2)
s4 = Student("Tony", 1)
s5 = Student("Jerry", 5)
预期结果:
result = ["Tony", "Sophie", "Bob", "Andy", "Jerry"]
我不确定是否需要创建对象数组或放置排序函数的位置。
有什么想法吗?
将其转换成对数组,然后应用合并排序并将数组作为参数传递
def merge(arr, l, m, r): n1 = m - l + 1 n2 = r- m
# create temp arrays
L = [0] * (n1)
R = [0] * (n2)
# Copy data to temp arrays L[] and R[]
for i in range(0 , n1):
L[i] = arr[l + i]
for j in range(0 , n2):
R[j] = arr[m + 1 + j]
# Merge the temp arrays back into arr[l..r]
i = 0 # Initial index of first subarray
j = 0 # Initial index of second subarray
k = l # Initial index of merged subarray
enter code here
while i < n1 and j < n2 :
if L[i][1] <= R[j][1]: // here you need to compare id
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Copy the remaining elements of L[], if there
# are any
while i < n1:
arr[k] = L[i]
i += 1
k += 1
# Copy the remaining elements of R[], if there
# are any
while j < n2:
arr[k] = R[j]
j += 1
k += 1
def mergeSort(arr,l,r):
if l < r:
# Same as (l+r)//2, but avoids overflow for
# large l and h
m = (l+(r-1))//2
# Sort first and second halves
mergeSort(arr, l, m)
mergeSort(arr, m+1, r)
merge(arr, l, m, r)
arr = [( "sdas",1), ( "asd3",3), ( "asd1",2)]
n = len(arr)
print ("Given array is")
for i in range(n):
print ("%s" %arr[i][0]),
mergeSort(arr,0,n-1)
print ("\n\nSorted array is")
for i in range(n):
print (arr[i]),
#or you can use predefined sort function like this
def sortSecond(val):
return val[1]
arr = [( "sdas",1), ( "asd3",3), ( "asd1",2)]
# sorts the array in ascending according to
# second element
arr.sort(key = sortSecond)
print(arr)
您可以使用 is 创建学生列表。 假设你将自己编写归并排序函数,现在你有一个这样的函数定义:
def merge_sort(students_array)
现在应该以特定方式实现此功能,您可以根据 id 比较元素。
为此,您有几种选择:
- 有一个__eq__(函数支持a == b),__le__(函数支持a <= b),__ge__(函数支持a >= b ), __gt__ (函数支持a > b), __lt__ (函数支持a < b)。您可以只使用“<、>、=”IMO,但这些是您要考虑的功能。 Student 需要实现它们,然后可以通过比较他们的 ID 来比较两个学生 - 这可能是最 pythonic 的方法:
示例实现:
class Student:
def __init__(self, name, id):
self.name = name
self.id = id
def __eq__(self, other):
return self.id == other.id
def __gt__(self, other):
return self.id > other.id
def __lt__(self, other):
return self.id < other.id
...
# getting a list of Students, not only ids
def merge_sort(students_array):
...
# example compare in the sort function
# this will be translated to: students_array[i].__gt__(students_array[j])
# which will return: students_array[i] > students_array[j]
if students_array[i] > students_array[j]:
# do something
...
def main():
s1 = Student("Andy", 4)
s2 = Student("Bob", 3)
s3 = Student("Sophie", 2)
s4 = Student("Tony", 1)
s5 = Student("Jerry", 5)
students_array = [s1, s2, s3, s4, s5]
merge_sort(students_array)
关于 python dunder 函数的良好文档: https://docs.python.org/3/library/operator.html
希望有所帮助