我一直在尝试修改打印到控制台的普通矩形2d数组，以显示其对角线以及其他字符。例如，我当前具有2d数组的矩形的代码是：
import pandas as pd
import numpy as np
df = pd.read_csv('whisky_analysis.csv')
df.head()
#get rid of the spaces!
df.rename(columns={'Meta Critic':'Score','Super Cluster':'Super_Cluster'}, inplace=True)
# replace special character $ with £
df.Cost = df.Cost.str.replace('$','£')
df.shape
df.isna().sum()
# Drop the 'Whisky', 'STDEV' and # columns
# Drop 'Cluster' and 'Super_Cluster' due to large numbers of NaN
df = df.drop(['Whisky', 'STDEV', '#', 'Super_Cluster', 'Cluster'], axis=1)
# drop any remaining rows with NaNs
df = df.dropna()
#Use one hot encoding to vectorise the categorical variables - we will use pandas' get_dummies
df1 = pd.get_dummies(df, columns = ['Cost', 'Type', 'Class','Country'])
# get target and feature vector/matrix
y = df1.pop('Score')
y.shape
X = df1
X.shape
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 2)
from sklearn.linear_model import LinearRegression
reg = LinearRegression().fit(X_train, y_train)
preds = reg.predict(X_test)
preds
# evaluating performance
from sklearn.metrics import mean_absolute_error
mean_absolute_error(y_test, preds)
from sklearn.metrics import r2_score
r2_score(y_test, preds)
并返回：
import java.util.Scanner;
class RecArray {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Height: ");
int height = scanner.nextInt();
System.out.print("Width: ");
int width = scanner.nextInt();
char[][] square = new char[height][width];
String line;
// fill the array
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
square[i][j] = 'o';
}
}
// print the array
for (int i = 0; i < height; i++) {
line = "";
for (int j = 0; j < width; j++) {
line += square[i][j];
}
System.out.println(line);
}
}
}
我希望对角线代码返回：
Height: 10
Width: 10
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
我当前的代码是：
Height: 5
Width: 7
xooooox
oxoooxo
ooxxxoo
oxoooxo
xooooox
这将返回：
import java.util.Scanner;
class RecArrayDiag {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Height: ");
int height = scanner.nextInt();
System.out.print("Width: ");
int width = scanner.nextInt();
char[][] square = new char[height][width];
boolean bool1 = true;
boolean bool2 = true;
boolean bool3 = true;
boolean bool4 = true;
String line;
int x = 0;
for (int i = 0; i < height; i++) {
for (int j = width-1; j >= 0; j--) {
if (i % 2 == 0 ? ((i == height/2)) : ((i == height-1/2))) {
bool1 = false;
}
if (j % 2 == 0 ? ((j == width/2)) : ((j == width-1/2))) {
bool2 = false;
}
if ((((i == j) && bool1 && bool2) || (i == (height - (j+1))) || (j == (width - (i+1))) || ((j == width-1) && bool3) || ((i == height-1) && bool4) || (j == width-1) && (i == height-1))) {
square[i][j] = 'x';
//x++;
} else {
square[i][j] = 'o';
}
if ((j == width-1)) {
bool3 = false;
}
if ((i == height-1)) {
bool4 = false;
}
}
x++;
}
// print the array
for (int i = 0; i < height; i++) {
line = "";
for (int j = 0; j < width; j++) {
line += square[i][j];
}
System.out.println(line);
}
}
}
请帮助我解决此问题，并先谢谢您。
答案 0 :(得分：2)
这是一种实现方法，可重用的方法将应用于矩形的各种操作分开。
public static void printRectangleWithDiagonals(int width, int height) {
char[][] rectangle = new char[height][width];
fill(rectangle, 'o');
drawDiagonals(rectangle, 'x');
print(rectangle);
}
private static void fill(char[][] rectangle, char ch) {
for (char[] line : rectangle)
for (int i = 0; i < line.length; i++)
line[i] = ch;
}
private static void drawDiagonals(char[][] rectangle, char ch) {
int bottom = rectangle.length - 1, right = rectangle[0].length - 1;
if (right > bottom) {
for (int x = 0; x <= right; x++) {
int y = (x * bottom + right / 2) / right;
rectangle[y][x] = ch;
rectangle[bottom - y][x] = ch;
}
} else {
for (int y = 0; y <= bottom; y++) {
int x = (y * right + bottom / 2) / bottom;
rectangle[y][x] = ch;
rectangle[y][right - x] = ch;
}
}
}
private static void print(char[][] rectangle) {
for (char[] line : rectangle)
System.out.println(line);
}
测试
printRectangleWithDiagonals(7, 7);
System.out.println();
printRectangleWithDiagonals(10, 4);
System.out.println();
printRectangleWithDiagonals(5, 9);
输出
xooooox
oxoooxo
ooxoxoo
oooxooo
ooxoxoo
oxoooxo
xooooox
xxooooooxx
ooxxxxxxoo
ooxxxxxxoo
xxooooooxx
xooox
oxoxo
oxoxo
ooxoo
ooxoo
oxoxo
oxoxo
xooox
xooox
答案 1 :(得分：2)
据我了解，您想显示某种十字形。 您要处理矩阵不是正方形的情况。
这意味着您可以直接从所有角转到中心点，如果一个轴到达数组的中间，则只需停止计数器并继续第二个参数即可。
类似的东西（只是伪代码）：
//create square with "o" everywhere then overwrite
int i = 0;
int j = 0;
while(i < height/2 || j < width/2){
//go from all corners towards the middle
if (i == j){
square[i][j] = "x";
square[i][width - j+1] = "x";
square[height - i+1][j] = "x";
square[height - i+1][width - j+1] = "x";
} else if (i < height/2) { //i is in middle of array
square[i][j] = "x";
square[i][width - j+1] = "x";
} else { //j is is in middle of array
square[i][j] = "x";
square[height - i+1][j] = "x";
}
//as long i and j did not reach the center add 1
if (i < width/2) { i++ }
if (j < height/2) { j++ }
}
希望这会有所帮助。 一般来说，我建议将您的问题分成不同的部分。
我可以在您的解决方案中看到逻辑，但请尝试使其保持简单。 查找只要条件为真的规则。 （在这种情况下：只要您不在任何数组中间） 然后尝试为不正确的情况找到解决方案。 （例如，如果我到达数组的中间但j没有到达数组的中间会发生什么）
就像您可以拆分代码并使其更易于阅读/维护一样。
在大多数情况下，如果您有大量的if else语句，则很有可能将它们重写为较小的部分。