Python 导入一个文本 识别文本内的数 然后做出成绩五级比较 输出每个的数量

可以的，您可以按照以下步骤实现：

1. 打开文本文件并读取文件内容

with open('filename.txt', 'r') as f:
    content = f.read()

2. 使用正则表达式提取文本中的数值

import re

numbers = re.findall(r'\d+', content)

3. 将提取出的数值转换为整数并根据分数划分等级

scores = []
for num in numbers:
    score = int(num)
    if score >= 90:
        scores.append('A')
    elif score >= 80:
        scores.append('B')
    elif score >= 70:
        scores.append('C')
    elif score >= 60:
        scores.append('D')
    else:
        scores.append('E')

4. 统计每个等级的数量并输出

from collections import Counter

score_count = Counter(scores)
for score, count in score_count.items():
    print(f"{score}: {count}")

完整代码如下：

import re
from collections import Counter

with open('filename.txt', 'r') as f:
    content = f.read()

numbers = re.findall(r'\d+', content)

scores = []
for num in numbers:
    score = int(num)
    if score >= 90:
        scores.append('A')
    elif score >= 80:
        scores.append('B')
    elif score >= 70:
        scores.append('C')
    elif score >= 60:
        scores.append('D')
    else:
        scores.append('E')

score_count = Counter(scores)
for score, count in score_count.items():
    print(f"{score}: {count}")