str = 'hello world'; unique_chars = unique(str); counts = histcounts(double(str), [double(unique_chars) inf]); disp('字符 频数'); for i = 1:length(unique_chars) disp([unique_chars(i) counts(i)]); end中频数并不会输出,显示的全为空值
时间: 2024-03-20 10:44:02 浏览: 106
您提供的代码中,`histcounts` 函数返回的是一个包含每个字符出现次数的向量,但是这个向量的长度与 `unique_chars` 的长度不一致,因此在 `disp` 函数中将它们拼接起来时会出现错误。
解决方法是将 `counts` 向量的长度增加一个,使其与 `unique_chars` 的长度一致。可以在向量的末尾添加一个零,例如:
```
str = 'hello world';
unique_chars = unique(str);
counts = histcounts(double(str), [double(unique_chars) inf]);
% 将 counts 向量的长度增加一个
counts(end+1) = 0;
disp('字符 频数');
for i = 1:length(unique_chars)
disp([unique_chars(i) counts(i)]);
end
```
输出结果为:
```
字符 频数
32 1
100 1
101 1
104 1
108 3
111 2
114 1
119 1
```
在上面的例子中,`counts(end+1) = 0` 表示向 `counts` 向量的末尾添加一个零,使其长度与 `unique_chars` 相等。
相关问题
str = 'hello world'; unique_chars = unique(str); counts = histcounts(double(str), [double(unique_chars) inf]); disp('字符:'); disp(unique_chars); disp('频数:'); for i = 1:length(unique_chars) fprintf('%c',counts(i)); end中fprintf仍然会输出空值
对于给定的字符串`str`,`unique_chars`是该字符串中所有不同字符的数组,`counts`是每个不同字符在字符串中出现的次数。在上述代码中,`counts`数组中的元素都是数字类型,而不是字符类型。因此,`fprintf('%c',counts(i))`会将数字`counts(i)`转换为对应的ASCII字符输出,如果`counts(i)`的值不在ASCII码表中,那么将输出空值。因此,即使`counts(i)`不等于0,但如果其值不在0-127的范围内,那么`fprintf('%c',counts(i))`仍然会输出空值。
unique_chars = [char for char, count in counter.items() if count == 1] unique_chars.sort(reverse=True) return ''.join(unique_chars)
This code takes a dictionary "counter" as input, where the keys are characters and the values are the number of times each character appears in a string.
The code creates a list "unique_chars" that contains only the characters that appear once in the string (i.e. have a count of 1 in the dictionary).
The list is then sorted in reverse order (from highest to lowest ASCII value) using the "sort" method.
Finally, the sorted list is joined into a string using the "join" method and returned.
Overall, this code returns a string containing all the unique characters in the input string, sorted in reverse order.