filename = "score.csv"
scores = {"程序设计": [], "体育": [], "经济学": []}
try:
with open(filename, "r") as file:
reader = csv.reader(file)
next(reader) # 跳过文件的第一行(表头)
for row in reader:
if len(row) == 4: # 确保每行都有三个成绩
_, score1, score2, score3 = row
scores["程序设计"].append(int(score1))
scores["体育"].append(int(score2))
scores["经济学"].append(int(score3))
for subject, subject_scores in scores.items():
average_score = sum(subject_scores) / len(subject_scores)
max_score = max(subject_scores)
min_score = min(subject_scores)
print(f"科目: {subject}")
print(f"平均分: {average_score:.2f}")
print(f"最高分: {max_score}")
print(f"最低分: {min_score}")
print()
直接用pandas去取csv文件的方式是最简单的,通过列索引就可以把数据给过滤出来,然后转化成列表,然后再求平均分、最高分、最低分。
import pandas as pd
# scores = {'程序设计': [], '体育': [], '经济学': []}
scores = {}
df = pd.read_csv('score.csv', encoding='utf-8') # 如果字符编码gbk的,需要把 utf-8 改成 gbk
scores['程序设计'] = df['程序设计'].to_list()
scores['体育'] = df['体育'].to_list()
scores['经济学'] = df['经济学'].to_list()
for k,v in scores.items():
print( '{}的平均分:{},最高分:{},最低分:{}'.format( k, round(sum(v) / len(v), 2), max(v), min(v) ) )
问题回答:
在定义字典变量时,需要注意的是确保键的唯一性,不能有重复的键;还需要确保对应的值具有相同的数据结构,例如在本问题里,需要确保三个键值对应的值都是空列表。
下面是具体的代码实现:
scores = {'程序设计': [], '体育': [], '经济学': []}
如果需要优化,可以考虑将字典的定义和初始化分开写,例如:
scores = {}
keys = ['程序设计', '体育', '经济学']
for key in keys:
scores[key] = []
这样可以提高代码的可读性和可维护性。
啥破代码有try没catch都不全的
先把代码好好的都粘代码块里粘全了,再把到底什么问题描述清楚