Pandas: 如何将两列连接为多行字符串？

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英文:

Pandas: How can I concatenate 2 columns as a multiline string?

问题

  1. 我有一个pandas数据框，想要使用多行字符串来连接它们。以下是我的pandas数据框，
  2. ```python
  3. pd.DataFrame([[1,"This is the desc of id 1"],[4,"This is the desc of id 2"]], columns=["id","desc"])

我想要的字符串如下，

  1. functionalRequirement 1{{
  2.   id: 1
  3.   text: This is the desc of id 1
  4.   risk: high
  5.   verifymethod: test
  6. }}

我尝试了下面的代码，但它没有工作。

  1. df['res'] = f"""
  2. functionalRequirement {df['id']} {{
  3.     id: {df['id']}
  4.     text: {df['desc']}
  5.     risk: high
  6.     verifymethod: test
  7. }} 
  8. """

我该如何做？

  2. <details>
  3. <summary>英文:</summary>
  5. I have a pandas dataframe and I want to concatenate them using multiline string. Below is my pandas dataframe,
  6. ```{python}
  7. pd.DataFrame([[1,&quot;This is the desc of id 1&quot;],[4,&quot;This is the desc of id 2&quot;]], columns=[&quot;id&quot;,&quot;desc&quot;])

I wanted string like below,

  1. functionalRequirement 1{{
  2.   id: 1
  3.   text: This is the desc of id 1
  4.   risk: high
  5.   verifymethod: test
  6. }}

I tried below code but it didnt work.

  1. df[&#39;res&#39;] = f&quot;&quot;&quot;
  2. functionaRequirement {df[&#39;id&#39;]} {{
  3.     id: {df[&#39;id&#39;]}
  4.     text: {df[&#39;desc&#39;]}
  5.     risk: high
  6.     verifymethod: test
  7. }} 
  8. &quot;&quot;&quot;

How can I do it?

答案1

得分: 2

Sure, here is the translated code portion:

  1. fstring = &quot;&quot;&quot;functionaRequirement {id} {{
  2.      id: {id}
  3.      text: {desc}
  4.      risk: high
  5.      verifymethod: test
  6.  }}&quot;&quot;&quot;
  8. df[&quot;res&quot;] = df.apply(lambda r: fstring.format(id=r.id, desc=r.desc), axis=1)
  10. &gt;&gt;&gt; print(df.res[0])
  11. functionaRequirement 1 {{
  12.     id: 1
  13.     text: 这是id 1的描述
  14.     risk: high
  15.     verifymethod: test
  16. }}

Please note that I have translated the text "This is the desc of id 1" to "这是id 1的描述" in the output.

英文:

Use string formatting and apply it over all rows

  2. fstring = &quot;&quot;&quot;functionaRequirement {id} {{
  3.      id: {id}
  4.      text: {desc}
  5.      risk: high
  6.      verifymethod: test
  7.  }}&quot;&quot;&quot;
  9. df[&quot;res&quot;] = df.apply(lambda r: fstring.format(id=r.id, desc=r.desc), axis=1)
  11. &gt;&gt;&gt; print(df.res[0])
  12. functionaRequirement 1 {
  13.     id: 1
  14.     text: This is the desc of id 1
  15.     risk: high
  16.     verifymethod: test
  17. }

答案2

得分: 1

你可以使用列表推导式：

  1. df['res'] = [f"""
  2. functionaRequirement {id} {{
  3.     id: {id}
  4.     text: {desc}
  5.     risk: high
  6.     verifymethod: test
  7. }} 
  8. """ for id, desc in zip(df['id'], df['desc'])]

或者如果性能很重要：

  1. ID = df['id'].astype(str)
  2. df['res'] = '\nfunctionaRequirement '+ID+' {\n    id: '+ID+'\n    text: '+df['desc']+'\n    risk: high\n    verifymethod: test\n}'

输出：

  1.    id                      desc                                                                                                                      res
  2. 0   1  This is the desc of id 1  \nfunctionaRequirement 1 {\n    id: 1\n    text: This is the desc of id 1\n    risk: high\n    verifymethod: test\n} 
  3. 1   4  This is the desc of id 2  \nfunctionaRequirement 4 {\n    id: 4\n    text: This is the desc of id 2\n    risk: high\n    verifymethod: test\n}
英文:

You could use a list comprehension:

  1. df[&#39;res&#39;] = [f&quot;&quot;&quot;
  2. functionaRequirement {id} {{
  3.     id: {id}
  4.     text: {desc}
  5.     risk: high
  6.     verifymethod: test
  7. }} 
  8. &quot;&quot;&quot; for id, desc in zip(df[&#39;id&#39;], df[&#39;desc&#39;])]

Or if performance matters:

  1. ID = df[&#39;id&#39;].astype(str)
  2. df[&#39;res&#39;] = &#39;\nfunctionaRequirement &#39;+ID+&#39; {\n    id: &#39;+ID+&#39;\n    text: &#39;+df[&#39;desc&#39;]+&#39;\n    risk: high\n    verifymethod: test\n}&#39;

Output:

  1.    id                      desc                                                                                                                      res
  2. 0   1  This is the desc of id 1  \nfunctionaRequirement 1 {\n    id: 1\n    text: This is the desc of id 1\n    risk: high\n    verifymethod: test\n} \n
  3. 1   4  This is the desc of id 2  \nfunctionaRequirement 4 {\n    id: 4\n    text: This is the desc of id 2\n    risk: high\n    verifymethod: test\n} \n

答案3

得分: 0

你可以使用 string 模板与 to_dict 导出你的数据框作为字典：

  1. import string
  3. TEMPLATE = string.Template("""\
  4. functionaRequirement ${id} {{
  5.     id: ${id}
  6.     text: ${desc}
  7.     risk: high
  8.     verifymethod: test
  9. }}""")
  11. df['res'] = list(map(TEMPLATE.substitute, df.to_dict('records')))
  12. # 或者
  13. df['res'] = [TEMPLATE.substitute(rec) for rec in df.to_dict('records')]

输出:

  1. >>> df
  2.    id                      desc                                                res
  3. 0   1  This is the desc of id 1  functionaRequirement 1 {{
  4.     id: 1
  5.     text: This is the desc of id 1
  6.     risk: high
  7.     verifymethod: test
  8. }}
  9. 1   4  This is the desc of id 2  functionaRequirement 4 {{
  10.     id: 4
  11.     text: This is the desc of id 2
  12.     risk: high
  13.     verifymethod: test
  14. }}
  16. >>> print(df.loc[0, 'res'])
  17. functionaRequirement 1 {{
  18.     id: 1
  19.     text: This is the desc of id 1
  20.     risk: high
  21.     verifymethod: test
  22. }}
英文:

You can use string template withT to_dict to export your dataframe as dict:

  1. import string
  3. TEMPLATE = string.Template(&quot;&quot;&quot;\
  4. functionaRequirement ${id} {{
  5.     id: ${id}
  6.     text: ${desc}
  7.     risk: high
  8.     verifymethod: test
  9. }}&quot;&quot;&quot;)
  11. df[&#39;res&#39;] = list(map(TEMPLATE.substitute, df.to_dict(&#39;records&#39;)))
  12. # OR
  13. df[&#39;res&#39;] = [TEMPLATE.substitute(rec) for rec in df.to_dict(&#39;records&#39;)]

Output:

  1. &gt;&gt;&gt; df
  2.    id                      desc                                                res
  3. 0   1  This is the desc of id 1  functionaRequirement 1 {{\n    id: 1\n    text...
  4. 1   4  This is the desc of id 2  functionaRequirement 4 {{\n    id: 4\n    text...
  6. &gt;&gt;&gt; print(df.loc[0, &#39;res&#39;])
  7. functionaRequirement 1 {{
  8.     id: 1
  9.     text: This is the desc of id 1
  10.     risk: high
  11.     verifymethod: test
  12. }}

huangapple
  • 本文由 发表于 2023年7月6日 18:14:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/76627765.html
  • pandas
  • python
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