我尝试使用另一个矩阵实例化coo_matrix。当我尝试打印(1, 9) 1.0
(1, 10) 1.0
(1, 11) 1.0
(1, 25) 1.0
(1, 47) 1.0
(2, 1) 1.0
(2, 7) 1.0
(2, 11) 3.0
(2, 12) 1.0
(2, 13) 1.0
(2, 15) 2.0
(2, 19) 1.0
(2, 42) 1.0
(3, 0) 1.0
(4, 20) 1.0
(4, 22) 1.0
(4, 24) 1.0
: :
(45, 0) 1.0
(45, 7) 1.0
(45, 14) 2.0
(45, 20) 1.0
(45, 26) 1.0
(45, 38) 1.0
(45, 40) 1.0
(46, 11) 1.0
(46, 19) 1.0
(46, 36) 1.0
(46, 41) 1.0
(46, 47) 1.0
时，输出为：
coo_matrix如何打印完整的set_printoptions(threshold = 'nan')？我尝试使用import java.util.*;
class bday {
public static void main(String[] args){
Scanner yearfinder = new Scanner(System.in);
System.out.println("Please type any year");
int year = yearfinder.nextInt();
Scanner monthfinder = new Scanner(System.in);
System.out.println("Please type any month");
String textmonth = monthfinder.nextLine();
int month;
switch(textmonth) {
case ("january"):
month = 1;
break;
case ("february"):
month = 2;
break;
case ("march"):
month = 3;
break;
case ("april"):
month = 4;
break;
case ("may"):
month = 5;
break;
case ("june"):
month = 6;
break;
case ("july"):
month = 7;
break;
case ("august"):
month = 8;
break;
case ("september"):
month = 9;
break;
case ("october"):
month = 10;
break;
case ("november"):
month = 11;
break;
case ("december"):
month = 12;
break;
default:
System.out.println("The month you input was invalid");
}
Scanner datefinder = new Scanner(System.in);
System.out.println("Please type any day");
int date = datefinder.nextInt();
System.out.println("The date you gave was " + date + "/" + month + "/" + year);
}
}
//Jan 1 1900 was a monday.
，但它并没有解决这个问题。
答案 0 :(得分：1)
您可以使用.todense()方法将稀疏矩阵转换为密集矩阵：
print(my_coo_matrix.todense())
编辑：您的问题听起来也像是要打印零值元素，但是如果您只想打印非零元素，则可以手动迭代矩阵：
for row, col, value in zip(my_coo_matrix.row, my_coo_matrix.col, my_coo_matrix.data):
print "({0}, {1}) {2}".format(row, col, value)
答案 1 :(得分：0)
我发现使用spy（）直观地表示它很有帮助。
spy(coo_matrix)
答案 2 :(得分：0)
（对于非常大的 coo_matrix）导出到 csv，然后将其保存为 xlsx 可能会有所帮助。
示例：
import numpy
numpy.savetxt('coo_matrix.csv', coo_matrix.todense(), delimiter = ',')