如何加速自定义函数

huangapple
go评论94阅读模式
英文:

How to speed up custom function

问题

以下是已翻译的部分：

  1. def keep_inum(row):
  2.     if len(row) != 0:
  3.         if int(row['inum']) in list1:
  4.             if row['DESC_1'] == 1:
  5.                 return row['recs']
  6.             else:
  7.                 return ''
  8.         elif int(row['inum']) in list2:
  9.             if row['DESC_1'] == 2:
  10.                 return row['recs']
  11.             else:
  12.                 return ''
  13.         elif int(row['inum']) in list3:
  14.             if row['DESC_1'] == 3:
  15.                 return row['recs']
  16.             else:
  17.                 return ''
  18.         else:
  19.             return row['recs']
  20.     else:
  21.         pass

应用函数到数据框（DF）：

  1. df['recs'] = df.apply(keep_inum, axis=1)
英文:

How to speed up my custom function?

I have three list of numbers :

list1
list2
list3

And Pandas Dataframe like this:

id inum DESC_1 recs
id1 inum1 1 recs1
id2 inum2 2 recs2
id3 inum3 3 recs3

And my custom function:

  1. def keep_inum(row):
  2.     if len(row) != 0:
  3.         if int(row['inum']) in list1:
  4.             if row['DESC_1'] == 1:
  5.                 return row['recs']
  6.             else:
  7.                 return ''
  8.         elif int(row['inum']) in list2:
  9.             if row['DESC_1'] == 2:
  10.                 return row['recs']
  11.             else:
  12.                 return ''
  13.         elif int(row['inum']) in list3:
  14.             if row['DESC_1'] == 3:
  15.                 return row['recs']
  16.             else:
  17.                 return ''
  18.         else:
  19.             return row['recs']
  20.     else:
  21.         pass

Apply func to DF:

  1. df['recs'] = df.apply(keep_inum, axis = 1)

答案1

得分: 1

不使用自定义函数：

  1. import pandas as pd
  3. df = pd.DataFrame(
  4.     {
  5.         "id": ["id1", "id2", "id3", "id4"],
  6.         "inum": ["111", "222", "333", "331"],
  7.         "DESC_1": [1, 4, 3, 3],
  8.         "recs": ["recs1", "recs2", "recs3", "yes"],
  9.     }
  10. )
  12. print(df)
  13. print("---")
  15. list1 = [111]
  16. list2 = [222]
  17. list3 = [333, 331]
  19. # 一次性将inum转换为整数
  20. df["inum_int"] = df["inum"].astype(int)
  21. # 清空不匹配desc的recs
  22. df.loc[df["inum_int"].isin(list1) & ~(df["DESC_1"] == 1), "recs"] = ""
  23. df.loc[df["inum_int"].isin(list2) & ~(df["DESC_1"] == 2), "recs"] = ""
  24. df.loc[df["inum_int"].isin(list3) & ~(df["DESC_1"] == 3), "recs"] = ""
  25. df.drop(columns=["inum_int"], inplace=True)
  26. print(df)
英文:

By not using a custom function at all:

  1. import pandas as pd
  3. df = pd.DataFrame(
  4.     {
  5.         "id": ["id1", "id2", "id3", "id4"],
  6.         "inum": ["111", "222", "333", "331"],
  7.         "DESC_1": [1, 4, 3, 3],
  8.         "recs": ["recs1", "recs2", "recs3", "yes"],
  9.     }
  10. )
  12. print(df)
  13. print("---")
  15. list1 = [111]
  16. list2 = [222]
  17. list3 = [333, 331]
  19. # Cast inum to int in one go
  20. df["inum_int"] = df["inum"].astype(int)
  21. # Empty the recs where inum doesn't match desc
  22. df.loc[df["inum_int"].isin(list1) & ~(df["DESC_1"] == 1), "recs"] = ""
  23. df.loc[df["inum_int"].isin(list2) & ~(df["DESC_1"] == 2), "recs"] = ""
  24. df.loc[df["inum_int"].isin(list3) & ~(df["DESC_1"] == 3), "recs"] = ""
  25. df.drop(columns=["inum_int"], inplace=True)
  26. print(df)

This outputs

  1.     id inum  DESC_1   recs
  2. 0  id1  111       1  recs1
  3. 1  id2  222       4  recs2
  4. 2  id3  333       3  recs3
  5. 3  id4  331       3    yes
  6. ---
  7.     id inum  DESC_1   recs
  8. 0  id1  111       1  recs1
  9. 1  id2  222       4       
  10. 2  id3  333       3  recs3
  11. 3  id4  331       3    yes

huangapple
  • 本文由 发表于 2023年2月6日 18:57:14
  • 转载请务必保留本文链接：https://go.coder-hub.com/75360409.html
  • function
  • pandas
  • python
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