修正下列代码 def classification_report_to_dict(report): lines = report.split('\n') lines = [line.strip() for line in lines if line.strip()] classes = [] class_dict = {} for line in lines[1:]: t = line.split() class_name = "" if len(t) == 1: class_name = t[0] classes.append(class_name) class_dict[class_name] = {} else: class_dict[class_name]['precision'] = float(t[0]) class_dict[class_name]['recall'] = float(t[1]) class_dict[class_name]['f1-score'] = float(t[2]) class_dict[class_name]['support'] = int(t[3]) macro_avg = lines[-3].split() micro_avg = lines[-2].split() class_dict['macro avg'] = {'precision': float(macro_avg[1]), 'recall': float(macro_avg[2]), 'f1-score': float(macro_avg[3]), 'support': int(macro_avg[4])} class_dict['micro avg'] = {'precision': float(micro_avg[1]), 'recall': float(micro_avg[2]), 'f1-score': float(micro_avg[3]), 'support': int(micro_avg[4])} return class_dict
时间: 2023-10-30 16:04:47 浏览: 34
这段代码用于将文本形式的分类器评价报告转换为 Python 字典格式。我看到这段代码实际上已经是一个完整的函数了,不过缺少了函数名。
如果你需要给这个函数命名,可以将 `def` 关键字之前的 `classification_report_to_dict(report):` 修改为你想要的函数名,例如 `def parse_classification_report(report):`。
另外,这个函数的实现是正确的,没有明显的错误。如果你有任何具体的问题或需要进一步的帮助,请告诉我。
相关问题
def classification_report_to_dict(report): lines = report.split('\n') lines = [line.strip() for line in lines if line.strip()] classes = [] class_dict = {} for line in lines[1:]: t = line.split() if len(t) == 1: class_name = t[0] classes.append(class_name) class_dict[class_name] = {} else: class_dict[class_name]['precision'] = float(t[0]) class_dict[class_name]['recall'] = float(t[1]) class_dict[class_name]['f1-score'] = float(t[2]) class_dict[class_name]['support'] = int(t[3]) macro_avg = lines[-3].split() micro_avg = lines[-2].split() class_dict['macro avg'] = {'precision': float(macro_avg[1]), 'recall': float(macro_avg[2]), 'f1-score': float(macro_avg[3]), 'support': int(macro_avg[4])} class_dict['micro avg'] = {'precision': float(micro_avg[1]), 'recall': float(micro_avg[2]), 'f1-score': float(micro_avg[3]), 'support': int(micro_avg[4])} return class_dict def classification_report_from_dict(report_dict): classes = list(report_dict.keys()) classes.remove('macro avg') classes.remove('micro avg') lines = [' precision recall f1-score support\n\n'] for class_name in classes: line = f"{class_name.ljust(15)}{report_dict[class_name]['precision']:.2f} {report_dict[class_name]['recall']:.2f} {report_dict[class_name]['f1-score']:.2f} {report_dict[class_name]['support']:5d}\n" lines.append(line) lines.append('\n') macro_avg = report_dict['macro avg'] line = f"{'macro avg'.ljust(15)}{macro_avg['precision']:.2f} {macro_avg['recall']:.2f} {macro_avg['f1-score']:.2f} {macro_avg['support']:5d}\n" lines.append(line) micro_avg = report_dict['micro avg'] line = f"{'micro avg'.ljust(15)}{micro_avg['precision']:.2f} {micro_avg['recall']:.2f} {micro_avg['f1-score']:.2f} {micro_avg['support']:5d}\n" lines.append(line) report_str = ''.join(lines) return report_str for i, report in enumerate(report): report_dict[f'report_{i + 1}'] = classification_report_to_dict(report) report_df = pd.DataFrame.from_dict(report_dict, orient='index') avg_report_dict = report_df.mean().to_dict() avg_report_str = classification_report_from_dict(avg_report_dict) print(avg_report_str)出现local variable 'class_name' referenced before assignment怎么解决
这个错误通常是因为在使用变量之前没有对其进行初始化或赋值,或者在使用变量之前使用了删除、重命名或声明之前的变量名。在你提供的代码中,我无法确定错误具体出现在哪一行,但是最可能的原因是在 `classification_report_from_dict` 函数中的 `classes` 列表为空,导致 `class_name` 变量没有被初始化。你可以在调用 `classification_report_from_dict` 函数之前,检查 `report_dict` 是否包含正确的数据。你也可以在 `classification_report_from_dict` 函数中添加一些错误处理机制,以避免出现未定义变量的情况。
metrics.classification_report
The `metrics.classification_report` function is a method used to generate a text report showing the main classification metrics of a classification model. The report includes precision, recall, F1 score and support for each class.
The function takes in four arguments:
- `y_true`: the true labels of the data
- `y_pred`: the predicted labels of the data
- `target_names`: the names of the classes (optional)
- `output_dict`: whether to return an output dictionary or not (optional, default is False)
Here's an example usage:
```
from sklearn.metrics import classification_report
# true labels
y_true = [0, 1, 2, 0, 1, 2]
# predicted labels
y_pred = [0, 2, 1, 0, 0, 1]
# class names
target_names = ['class 0', 'class 1', 'class 2']
# generate report
print(classification_report(y_true, y_pred, target_names=target_names))
```
The output will be something like:
```
precision recall f1-score support
class 0 0.67 1.00 0.80 2
class 1 0.00 0.00 0.00 2
class 2 1.00 0.50 0.67 2
accuracy 0.50 6
macro avg 0.56 0.50 0.49 6
weighted avg 0.56 0.50 0.49 6
```
The report shows the precision, recall, F1 score and support for each class. It also includes an overall accuracy and macro and weighted averages of the metrics.