Python从列表转换为字符串后删除字符串中的字符。

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英文:

Python remove characters in string that was converted from list

问题

I have a list that is structured like so:

  1. [{'Id'}, {'IsDeleted'}, {'MasterRecordId'}, {'Name'}, {'Title'}, {'Company'}]

I want to turn it into a string that looks like this:

  1. Id, IsDeleted, MasterRecordId, Name, Title, Company

Here's the code I have so far:

  1. sf = Salesforce(instance_url='insideigt--tony.sandbox.my.salesforce.com/',
  2.                 username='igtdataimport@igt.com.tony', 
  3.                 password='TXRZ51rRe',
  4.                 security_token='2IagJSrI5H3zxeVjgqy1xAN7d',
  5.                 domain='test')
  7. def getValidFields(sObject):
  8.     fields = []
  10.     print(sObject)
  11.     schema = getattr(sf, sObject).describe()
  13.     for fieldDict in schema.get('fields', []):
  14.         fieldType = fieldDict.get('type')
  15.         if fieldType not in ['address', 'location'] and not fieldDict.get('compoundFieldName'):
  16.             fields.append(fieldDict.get('name'))
  18.     my_string = ', '.join(fields)
  20.     print(my_string)
  21.     return fields
  23. sObject = 'Lead'
  24. fields = getValidFields(sObject)

I hope this helps!

英文:

I have a list that is structured like so:

  1. [{'Id'}, {'IsDeleted'}, {'MasterRecordId'}, {'Name'}, {'Title'}, {'Company'}]

I want to turn it into a string that looks like this:

  1. Id, IsDeleted, MasterRecordId, Name, Title, Company

I've gone through the top answer here:

https://stackoverflow.com/questions/10017147/removing-a-list-of-characters-in-string

and a few others.

Here's the code I have so far:

  1.     sf = Salesforce(instance_url='insideigt--tony.sandbox.my.salesforce.com/',
  2.                 username='igtdataimport@igt.com.tony', 
  3.                 password='TXRZ51rRe',
  4.                 security_token='2IagJSrI5H3zxeVjgqy1xAN7d',
  5.                 domain='test')
  7. def getValidFeilds( sObject ):
  8.     fields = []
  10.     print(sObject)
  11.     schema = getattr( sf, sObject).describe()
  13.     for fieldDict in schema.get('fields', []):
  14.         fieldType = fieldDict.get('type')
  15.         if fieldType not in ['address', 'location'] and not fieldDict.get('compoundFieldName'):
  16.             fields.append({
  17.                 fieldDict.get('name')
  18.             })
  19.     #np.array(fields).flatten()
  20.     chars_to_remove = ["'", "{" ,"}" ]
  21.     my_string= ','.join(map(str, fields))
  22.     my_string.translate(str.maketrans({"'":None}))
  25.     
  26.     print(my_string)
  27.     return fields
  28. sObject = 'Lead'
  29. fields = getValidFeilds(sObject)

答案1

得分: 0

这是您的代码的更新版本，应该可以工作：

  1. def getValidFields(sObject):
  2.     fields = []
  4.     schema = getattr(sf, sObject).describe()
  6.     for fieldDict in schema.get('fields', []):
  7.         fieldType = fieldDict.get('type')
  8.         if fieldType not in ['address', 'location'] and not fieldDict.get('compoundFieldName'):
  9.             fields.append(fieldDict.get('name'))
  11.     # 使用 `join` 函数将字段列表连接成一个字符串，用逗号分隔
  12.     my_string = ', '.join(fields)
  14.     # 使用 `translate` 函数删除不想要的字符
  15.     # 这将从字符串中删除单引号、花括号和空格
  16.     my_string = my_string.translate(str.maketrans('', '', "'{} "))
  18.     print(my_string)
  19.     return fields
  21. sObject = 'Lead'
  22. fields = getValidFields(sObject)
英文:

Here's an updated version of your code that should work:

  1. def getValidFeilds( sObject ):
  2.     fields = []
  4.     schema = getattr( sf, sObject).describe()
  6.     for fieldDict in schema.get('fields', []):
  7.         fieldType = fieldDict.get('type')
  8.         if fieldType not in ['address', 'location'] and not fieldDict.get('compoundFieldName'):
  9.             fields.append(fieldDict.get('name'))
  11.     # Use the `join` function to join the list of fields into a single string,
  12.     # separated by commas
  13.     my_string = ', '.join(fields)
  15.     # Remove the characters that you don't want using the `translate` function
  16.     # This will remove single quotes, curly braces, and spaces from the string
  17.     my_string = my_string.translate(str.maketrans('', '', "'{} "))
  19.     print(my_string)
  20.     return fields
  22. sObject = 'Lead'
  23. fields = getValidFeilds(sObject)

答案2

得分: 0

我认为你示例中的花括号表示集合，所以你的列表实际上是一个集合列表。

这将为您提供一个只包含值的列表 - 如果您愿意，您可以将其改为列表理解而不是循环。

  1. list_of_sets = [{'Id'}, {'IsDeleted'}, {'MasterRecordId'}, {'Name'}, {'Title'}, {'Company'}]
  3. new_list = []
  4. for n in list_of_sets:
  5.     for y in n:
  6.         new_list.append(y)
  8. print(new_list)
英文:

I think the curly brackets in your example are representing sets, so your list is actually a list of sets.

This should give you a list of just the values - you could make it a list comprehension instead of a loop if you wanted.

  1. list_of_sets = [{'Id'}, {'IsDeleted'}, {'MasterRecordId'}, {'Name'}, {'Title'}, {'Company'}]
  3. new_list = []
  4. for n in list_of_sets:
  5.     for y in n:
  6.         new_list.append(y)
  8. print(new_list)

答案3

得分: 0

S = " ".join(word for word, *_ in L)

print(S)
Id IsDeleted MasterRecordId Name Title Company

英文:

you could use join with a comprehension that extracts the word from each set:

  1. L = [{'Id'}, {'IsDeleted'}, {'MasterRecordId'}, {'Name'}, 
  2.      {'Title'}, {'Company'}]
  4. S = " ".join( word for word,*_ in L )
  6. print(S)
  7. Id IsDeleted MasterRecordId Name Title Company

huangapple
  • 本文由 发表于 2023年3月7日 03:17:04
  • 转载请务必保留本文链接：https://go.coder-hub.com/75654942.html
  • python
  • python-3.x
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