[已解决]请大师帮做个键是6位数的字典，谢谢

futui

请大师帮做个字典，谢谢

1:000395/菲利华/9.38%/536.81/29,524.39。；2:00593/新雷能/8.82%/653.62/27,765.83。；10:603105/芯能科技/2.51%/535.72/7,891.21。；2:1593/新雷/8.82%/653.62/27,765.83。


怎么把上面做成如下字典：键是冒号和斜杠之间的6位数，值是第二个斜杠后的值

{‘000395’:9.38,‘603105’:2.51}

最佳答案

月排行榜 / 总排行榜

isdkz

2023-5-4 14:07:30

可以通过正则表达式和Python字典解析的方式实现，代码如下：

1. import re
   
2. 
   
3. text = "1:000395/菲利华/9.38%/536.81/29,524.39。；2:00593/新雷能/8.82%/653.62/27,765.83。；10:603105/芯能科技/2.51%/535.72/7,891.21。；2:1593/新雷/8.82%/653.62/27,765.83。"
   
4. 
   
5. pattern = re.compile(r'(\d{6})/[^/]*/(\d+\.\d+)%')
   
6. 
   
7. result = {m.group(1): float(m.group(2)) for m in pattern.finditer(text)}
   
8. 
   
9. print(result)
   

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运行上面的代码，将得到所需的字典结果：

1. {'000395': 9.38, '603105': 2.51}
   

复制代码

跳转到最佳答案楼层

isdkz

可以通过正则表达式和Python字典解析的方式实现，代码如下：

1. import re
   
2. 
   
3. text = "1:000395/菲利华/9.38%/536.81/29,524.39。；2:00593/新雷能/8.82%/653.62/27,765.83。；10:603105/芯能科技/2.51%/535.72/7,891.21。；2:1593/新雷/8.82%/653.62/27,765.83。"
   
4. 
   
5. pattern = re.compile(r'(\d{6})/[^/]*/(\d+\.\d+)%')
   
6. 
   
7. result = {m.group(1): float(m.group(2)) for m in pattern.finditer(text)}
   
8. 
   
9. print(result)
   

复制代码

运行上面的代码，将得到所需的字典结果：

1. {'000395': 9.38, '603105': 2.51}
   

复制代码

wp231957

普通思路  仍然可行：

1. s="1:300395/菲利华/9.38%/536.81/29,524.39。；2:300593/新雷能/8.82%/653.62/27,765.83。；10:60305/芯能科技/2.51%/535.72/7,891.21"
   
2. dic={}
   
3. for x in s.split("；"):
   
4.     m=x.split("/")
   
5.     if len(m[0].split(":")[1])==6:
   
6.          dic[m[0].split(":")[1]]=m[2][:-1]
   
7. print(dic)   

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