Nested for loop taking exact list items – Python

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英文:

Nested for loop taking exact list items - Python

问题

  1. <Partners>
  2.   <ExternalCode>aaa</ExternalCode>
  3.   <ShortName>ddd</ShortName>
  4.   <Name>ggg</Name>
  5.   <PartnerGroupCode>jjj</PartnerGroupCode>
  6. </Partners>
  7. <Partners>
  8.   <ExternalCode>aaa</ExternalCode>
  9.   <ShortName>ddd</ShortName>
  10.   <Name>ggg</Name>
  11.   <PartnerGroupCode>kkk</PartnerGroupCode>
  12. </Partners>
  13. ...
  14. (共计 9 个类似的块)
英文:

I am trying to write a code for building a XML file which has to have about 1000 lines of data about partners.

I want to take the loops to take for example from list_a first item and match it with first item of list_b. second with second and so on.

  1. list_a = ['aaa', 'bbb', 'ccc']
  2. list_b = ['ddd', 'eee', 'fff']
  3. list_c = ['ggg', 'hhh', 'iii']
  4. list_d = ['jjj', 'kkk', 'lll']

here is the code I have tried. Cannot find the right breaks from the loop so the code can give proper XML text.

  1. for item_a in list_a:  
  2.   print('<Partners>')
  3.   print('  <ExternalCode>', end='')
  4.   print(item_a, end='')
  5.   print('</ExternalCode>')
  6.   for item_b in list_b:
  7.     print('  <ShortName>', end='')
  8.     print(item_b, end='')
  9.     print('</ShortName>')
  10.     for item_c in list_c:
  11.       print('  <Name>', end='')
  12.       print(item_c, end='')
  13.       print('</Name>')
  14.       for item_d in list_d:
  15.         print('  <PartnerGroupCode>', end='')
  16.         print(item_d, end='')
  17.         print('</PartnerGroupCode>')
  18.         print('</Partners>')

The code has to give an XML text like this:

  1. <Partners>
  2.   <ExternalCode>String 33</ExternalCode>
  3.   <ShortName>String 34</ShortName>
  4.   <Name>String 35</Name>
  5.   <PartnerGroupCode>String 36</PartnerGroupCode>
  6. </Partners>
  7. <Partners>
  8.   <ExternalCode>String 37</ExternalCode>
  9.   <ShortName>String 38</ShortName>
  10.   <Name>String 39</Name>
  11.   <PartnerGroupCode>String 40</PartnerGroupCode>
  12. </Partners>

答案1

得分: 1

  1. 如果我理解正确，您想以以下方式将列表中的相应元素分组？
  3. Output
  4. <Partners>
  5.   <ExternalCode>aaa</ExternalCode>
  6.   <ShortName>ddd</ShortName>
  7.   <Name>ggg</Name>
  8.   <PartnerGroupCode>jjj</PartnerGroupCode>
  9. </Partners>
  10. <Partners>
  11.   <ExternalCode>bbb</ExternalCode>
  12.   <ShortName>eee</ShortName>
  13.   <Name>hhh</Name>
  14.   <PartnerGroupCode>kkk</PartnerGroupCode>
  15. </Partners>
  16. <Partners>
  17.   <ExternalCode>ccc</ExternalCode>
  18.   <ShortName>fff</ShortName>
  19.   <Name>iii</Name>
  20.   <PartnerGroupCode>lll</PartnerGroupCode>
  21. </Partners>
英文:

if my understanding is correct you would like to group all the corresponding elements in a list in the below manner ?

  1. list_a = ['aaa', 'bbb', 'ccc']
  2. list_b = ['ddd', 'eee', 'fff']
  3. list_c = ['ggg', 'hhh', 'iii']
  4. list_d = ['jjj', 'kkk', 'lll']
  6. for i in range(len(list_a)):
  7.     print('<Partners>')
  8.     print('  <ExternalCode>' + list_a[i] + '</ExternalCode>')
  9.     print('  <ShortName>' + list_b[i] + '</ShortName>')
  10.     print('  <Name>' + list_c[i] + '</Name>')
  11.     print('  <PartnerGroupCode>' + list_d[i] + '</PartnerGroupCode>')
  12.     print('</Partners>')

Output

  1. <Partners>
  2.   <ExternalCode>aaa</ExternalCode>
  3.   <ShortName>ddd</ShortName>
  4.   <Name>ggg</Name>
  5.   <PartnerGroupCode>jjj</PartnerGroupCode>
  6. </Partners>
  7. <Partners>
  8.   <ExternalCode>bbb</ExternalCode>
  9.   <ShortName>eee</ShortName>
  10.   <Name>hhh</Name>
  11.   <PartnerGroupCode>kkk</PartnerGroupCode>
  12. </Partners>
  13. <Partners>
  14.   <ExternalCode>ccc</ExternalCode>
  15.   <ShortName>fff</ShortName>
  16.   <Name>iii</Name>
  17.   <PartnerGroupCode>lll</PartnerGroupCode>
  18. </Partners>
  20. </details>
  24. # 答案2
  25. **得分**: 1
  27. Here is the translated code:
  29. ```python
  30. 另一种通用的从一组列表中获取第i个元素的方法是使用Python的zip函数。
  32. for a, b, c, d in zip(list_a, list_b, list_c, list_d):
  33.     print('<Partners>')
  34.     print(f'  <ExternalCode>{a}</ExternalCode>')
  35.     print(f'  <ShortName>{b}</ShortName>')
  36.     print(f'  <Name>{c}</Name>')
  37.     print(f'  <PartnerGroupCode>{d}</PartnerGroupCode>')
  38.     print('</Partners>')

Please note that I have translated the code portion as requested, and there are no additional contents or answers to translation questions.

英文:

Another general approach to taking the i-th element from a group of lists is the python zip function.

  1. for a, b, c, d in zip(list_a, list_b, list_c, list_d):
  2.     print(&#39;&lt;Partners&gt;&#39;)
  3.     print(f&#39;  &lt;ExternalCode&gt;{a}&lt;/ExternalCode&gt;&#39;)
  4.     print(f&#39;  &lt;ShortName&gt;{b}&lt;/ShortName&gt;&#39;)
  5.     print(f&#39;  &lt;Name&gt;{c}&lt;/Name&gt;&#39;)
  6.     print(f&#39;  &lt;PartnerGroupCode&gt;{d}&lt;/PartnerGroupCode&gt;&#39;)
  7.     print(&#39;&lt;/Partners&gt;&#39;)

huangapple
  • 本文由 发表于 2023年8月11日 01:27:08
  • 转载请务必保留本文链接：https://go.coder-hub.com/76878037.html
  • for-loop
  • nested-loops
  • python
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