[英]Using merge sort/quick sort to sort attribute of class objects in Python
我有這個學生 class
class Student:
def __init__(self, name, id):
self.name = name
self.id = id
我需要對學生 class 中的一些對象進行排序,特別是使用合並/快速排序對id
進行排序,預期結果是學生姓名的數組。 所以如果我有這些對象:
s1 = Student("Andy", 4)
s2 = Student("Bob", 3)
s3 = Student("Sophie", 2)
s4 = Student("Tony", 1)
s5 = Student("Jerry", 5)
和預期的結果:
result = ["Tony", "Sophie", "Bob", "Andy", "Jerry"]
我不確定我是否需要創建 object 數組或我將排序 function 放在哪里。
有什么想法嗎?
您可以使用 is 創建學生列表。 假設您將自己編寫合並排序 function,現在您有一個 function 定義,如下所示:
def merge_sort(students_array)
現在這個 function 應該以特定的方式實現,您可以根據 id 比較元素。
為此,您有幾個選擇:
示例實現:
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)
關於 python dunder 功能的良好文檔: https://docs.python.org/3/library/operator.ZFC35EZ883622
希望有幫助
將此轉換為對數組,然后應用合並排序並將數組作為參數傳遞
def 合並(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)
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