如何按某列对数据框进行排序？

huangapple
go评论119阅读模式
英文:

How can I sort a dataframe on some column?

问题

我想按照以下顺序对数据进行排序：（9, 10, 11, 12, 1, 2, 3），你能帮我做到吗？

英文:

I have the following code:

  1. import pandas as pd
  2. DF = pd.DataFrame({
  3.     "item_id": [1, 2, 3, 4, 5, 6, 7, 8, 9],
  4.     "customer": ["CG", "CGD", "CC", "CZ", "JD", "KD", "MM", "MK", "JJ"],
  5.     "costs": [150, 12, 78, 56, 5, 6, 4, 68, 69],
  6.     "games": [["Minecraft", "Uno"], ["WOW", "Minecraft"], ["WOW", "Minecraft"], ["WOW", "Minecraft", "The last of us"], ["Monopoly"], ["Monopoly", "Uno"], ["WOW", "Monopoly"], ["Uno", "SkipBo"], ["Schach", "Uno"]], 
  7.     "date": ["2022-01-18", "2022-03-12", "2022-02-26", "2021-11-12", "2021-10-09", "2021-11-26", "2021-10-18", "2021-12-22", "2021-09-06"]
  8. })
  10. DF["games"] = [i[-1:] for i in DF["games"]]
  11. DF["Day"] = pd.to_datetime(DF["date"]).dt.day
  12. DF["Month"] = pd.to_datetime(DF["date"]).dt.month
  13. DF["Year"] = pd.to_datetime(DF["date"]).dt.year
  14. DF #(look at pic_1)
  16. newDF_month = DF.drop(columns=["item_id", "customer", "games", "date", "Day", "Year"])
  17. newDF_month.groupby(["Month"]).sum()

But this presents the data ordered by month using the ordering (1,2,3,9,10,11,12):

  1. month | costs
  2. -------------
  3. 1  | 150
  4. 2  |  78
  5. 3  |  12
  6. 9  |  69
  7. 10  |   9
  8. 11  |  62
  9. 12  |  68

but I would like to have the ordering be (9, 10, 11, 12, 1, 2, 3) instead:

  1. month | costs
  2. -------------
  3. 9  |  69
  4. 10  |   9
  5. 11  |  62
  6. 12  |  68
  7. 1  | 150
  8. 2  |  78
  9. 3  |  12

How can I do that?

答案1

得分: 4

你可以对其移位后取模12来使用 sort_index：

  1. (newDF_month
  2.  .groupby("Month").sum()
  3.  .sort_index(key=lambda i: (i-9)%12)
  4. )

输出：

  1.        costs
  2. Month       
  3. 9         69
  4. 10         9
  5. 11        62
  6. 12        68
  7. 1        150
  8. 2         78
  9. 3         12
英文:

You can sort_index on its shifted value modulo 12:

  1. (newDF_month
  2. .groupby(["Month"]).sum()
  3. .sort_index(key=lambda i: (i-9)%12)
  4. )

Output:

  1.        costs
  2. Month       
  3. 9         69
  4. 10         9
  5. 11        62
  6. 12        68
  7. 1        150
  8. 2         78
  9. 3         12

答案2

得分: 1

另一个可能的解决方案：

  1. 新 = 新DF_月.groupby(["月"]).sum()
  2. m = 新.index >= 9
  3. pd.concat([新.loc[m, :], 新.loc[~m, :]])

输出：

  1.        费用
  2. 月       
  3. 9       69
  4. 10       9
  5. 11      62
  6. 12      68
  7. 1      150
  8. 2       78
  9. 3       12
英文:

Another possible solution:

  1. new = newDF_month.groupby(["Month"]).sum()
  2. m = new.index >= 9
  3. pd.concat([new.loc[m, :], new.loc[~m, :]])

Output:

  1.        costs
  2. Month       
  3. 9         69
  4. 10         9
  5. 11        62
  6. 12        68
  7. 1        150
  8. 2         78
  9. 3         12

huangapple
  • 本文由 发表于 2023年4月16日 23:38:39
  • 转载请务必保留本文链接：https://go.coder-hub.com/76028704.html
  • dataframe
  • pandas
  • python
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定