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法学编写 Python 程序来找出二进制矩阵中的重复行
给定一个包含 0 和 1 的二进制矩阵，我们的任务是找到重复行并打印出来。
Python 提供了用于此处的 Counter() 方法。
示例
Input: 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 Output: (1, 1, 1, 1) (0, 0, 0, 0)
算法
Step 1: Create a binary matrix, only 0 and 1 elements are present. Step 2: Which will have rows as key and it’s frequency as value. Lists are mutable so first, we will cast each row (list) into a tuple. Step 3: Create a dictionary using the counter method. Step 4: Now traverse the dictionary completely. Step 5: Print all rows which have frequency greater than 1.
样例代码
# Function to find duplicate rows in a binary matrix
from collections import Counter
def binarymatrix(A):
A = map(tuple,A)
dic = Counter(A)
print("Duplicate rows of Binary Matrix ::>")
for (i,j) in dic.items():
if j>1:
print (i)
# Driver program
if __name__ == "__main__":
A=[]
n=int(input("Enter n for n x n matrix : ")) #3 here
#use list for storing 2D array
#get the user input and store it in list (here IN : 1 to 9)
print("Enter the element ::>")
for i in range(n):
row=[] #temporary list to store the row
for j in range(n):
row.append(int(input())) #add the input to row list
A.append(row) #add the row to the list
print(A)
# [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
#Display the 2D array
print("Display Array In Matrix Form")
for i in range(n):
for j in range(n):
print(A[i][j], end=" ")
print()
binarymatrix(A)输出
Enter n for n x n matrix : 4 Enter the element ::> 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 [[1, 1, 1, 1], [0, 0, 0, 0], [1, 1, 1, 1], [0, 0, 0, 0]] Display Array In Matrix Form 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 Duplicate rows of Binary Matrix ::> (1, 1, 1, 1) (0, 0, 0, 0)
更新于： 23-6 月-2020
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