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])中“class_name”出错误怎么纠正python代码

这段 Python 代码中的 `class_name` 变量没有初始化，可能会导致 `NameError` 异常。您可以将 `class_name` 初始化为一个空字符串，如下所示：

class_name = ''
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])

这样会在第一次处理 `len(t) == 1` 的情况时，为 `class_name` 赋值，避免了 `NameError` 错误。

相关问题

lines = report.split('\n') lines = [line.strip() for line in lines if line.strip()] class_name = '' classes = [] class_dict = {} default_metrics = {'precision': 0.0, 'recall': 0.0, 'f1-score': 0.0, 'support': 0} for line in lines[1:]: t = line.split() if len(t) == 1: class_name = t[0] classes.append(class_name) class_dict[class_name] = default_metrics.copy() 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])总是出现KeyError: ''怎么修正python

要修正这个 KeyError: '' 错误，可以在处理第一行时检查其内容是否为空字符串。可以使用一个 if 语句来判断当前行是否为空字符串，如果是空字符串，则跳过当前循环。代码如下：

lines = report.split('\n')
lines = [line.strip() for line in lines if line.strip()]
class_name = ''
classes = []
class_dict = {}
default_metrics = {'precision': 0.0, 'recall': 0.0, 'f1-score': 0.0, 'support': 0}

for line in lines[1:]:
    t = line.split()
    if not t: # 如果当前行为空字符串，则跳过当前循环
        continue
    if len(t) == 1:
        class_name = t[0]
        classes.append(class_name)
        class_dict[class_name] = default_metrics.copy()
    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])

这样就能够避免因为空字符串导致的 KeyError: '' 错误。

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

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。