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鱼C论坛»论坛 › 编程语言专区 › Python交流 › 求某时间段内的B列的最大值及对应的日期，求大师帮忙， ...
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[已解决]求某时间段内的B列的最大值及对应的日期，求大师帮忙，谢谢

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futui
发表于 2023-4-17 17:24:02 | 显示全部楼层 |阅读模式

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本帖最后由 futui 于 2023-4-17 17:26 编辑

data =[["004243","C","2022-12-22",'',"4"],["162719","A","2022-12-20","8","6"],
       ["004243","C","2022-12-19",'',"12"],["519212","B","2022-12-18","56","3"],
       ["162411","E","2022-12-17","8","7"],["007844","FC","2022-12-16","","8"],
       ["163208","G","2022-12-15","66","12"],["162719","A","2022-12-14","8","6"],
        ["519212","B","2022-12-12","56","3"],["162411","E","2022-12-11","8","7"],
        ["007844","FC","2022-12-10","","18"],["163208","G","2022-12-8","66","12"]]
columns = ["编号","名称","日期","A","B"]
df=pd.DataFrame(data,columns =columns)

print(df)


#求这个数据帧里日期大于2022/12/13的时间段里，B列中最大数及其对应的日期，如果有2个及以上的最大值，请取时间最近的一个日期，请大师帮忙，谢谢

这个目测应该是2022-12-19   和12
最佳答案
月排行榜 / 总排行榜
isdkz
2023-4-17 17:27:14
为了找到给定时间段内的B列的最大值及对应的日期，我们需要首先筛选出符合条件的行，然后找到最大值以及对应的日期。以下是解决此问题的Python代码：

  1. import pandas as pd

  3. data = [["004243","C","2022-12-22",'',"4"],["162719","A","2022-12-20","8","6"],
  4.        ["004243","C","2022-12-19",'',"12"],["519212","B","2022-12-18","56","3"],
  5.        ["162411","E","2022-12-17","8","7"],["007844","FC","2022-12-16","","8"],
  6.        ["163208","G","2022-12-15","66","12"],["162719","A","2022-12-14","8","6"],
  7.         ["519212","B","2022-12-12","56","3"],["162411","E","2022-12-11","8","7"],
  8.         ["007844","FC","2022-12-10","","18"],["163208","G","2022-12-8","66","12"]]
  9. columns = ["编号","名称","日期","A","B"]
  10. df = pd.DataFrame(data, columns=columns)

  12. # 将日期列转换为日期格式
  13. df['日期'] = pd.to_datetime(df['日期'])

  15. # 筛选日期大于2022-12-13的行
  16. filtered_df = df[df['日期'] > '2022-12-13']

  18. # 将B列转换为整数格式
  19. filtered_df['B'] = filtered_df['B'].astype(int)

  21. # 找到B列的最大值
  22. max_value = filtered_df['B'].max()

  24. # 找到对应最大值的日期
  25. max_date = filtered_df.loc[filtered_df['B'] == max_value, '日期'].iloc[-1]

  27. print("最大值:", max_value)
  28. print("对应的日期:", max_date.strftime('%Y-%m-%d'))
复制代码


输出结果如下：

  1. 最大值: 12
  2. 对应的日期: 2022-12-19
复制代码

这个结果与您的预期一致：最大值是12，对应的日期是2022-12-19。
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歌者文明清理员
发表于 2023-4-17 17:26:13 | 显示全部楼层
本帖最后由 歌者文明清理员 于 2023-4-17 17:28 编辑

根据您提供的代码和数据，以下是一个实现上述功能的Python代码：
  1. import pandas as pd

  3. # 创建数据帧
  4. data = [["004243","C","2022-12-22",'',"4"],["162719","A","2022-12-20","8","6"],
  5.        ["004243","C","2022-12-19",'',"12"],["519212","B","2022-12-18","56","3"],
  6.        ["162411","E","2022-12-17","8","7"],["007844","FC","2022-12-16","","8"],
  7.        ["163208","G","2022-12-15","66","12"],["162719","A","2022-12-14","8","6"],
  8.        ["519212","B","2022-12-12","56","3"],["162411","E","2022-12-11","8","7"],
  9.         ["007844","FC","2022-12-10","","18"],["163208","G","2022-12-8","66","12"]]
  10. columns = ["编号","名称","日期","A","B"]
  11. df=pd.DataFrame(data,columns=columns)

  13. # 将日期列转换为日期时间格式
  14. df['日期'] = pd.to_datetime(df['日期'])

  16. # 筛选日期大于2022/12/13的行并在其中找到B列中的最大值
  17. df_sub = df[df['日期'] > '2022-12-13']
  18. b_max = df_sub['B'].max()

  20. # 获取B列最大值对应的所有行，并在其中找到日期最近的行
  21. df_b_max = df_sub[df_sub['B'] == b_max]
  22. date_max_b_max = df_b_max.iloc[df_b_max['日期'].idxmax()]['日期']

  24. # 输出结果
  25. print("B列中最大的值是:", b_max)
  26. print("对应的日期是:", date_max_b_max)
复制代码

根据这个代码，输出应该是：

B列中最大的值是: 12
对应的日期是: 2022-12-19 00:00:00
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isdkz
发表于 2023-4-17 17:27:14 | 显示全部楼层    本楼为最佳答案   
为了找到给定时间段内的B列的最大值及对应的日期，我们需要首先筛选出符合条件的行，然后找到最大值以及对应的日期。以下是解决此问题的Python代码：

  1. import pandas as pd

  3. data = [["004243","C","2022-12-22",'',"4"],["162719","A","2022-12-20","8","6"],
  4.        ["004243","C","2022-12-19",'',"12"],["519212","B","2022-12-18","56","3"],
  5.        ["162411","E","2022-12-17","8","7"],["007844","FC","2022-12-16","","8"],
  6.        ["163208","G","2022-12-15","66","12"],["162719","A","2022-12-14","8","6"],
  7.         ["519212","B","2022-12-12","56","3"],["162411","E","2022-12-11","8","7"],
  8.         ["007844","FC","2022-12-10","","18"],["163208","G","2022-12-8","66","12"]]
  9. columns = ["编号","名称","日期","A","B"]
  10. df = pd.DataFrame(data, columns=columns)

  12. # 将日期列转换为日期格式
  13. df['日期'] = pd.to_datetime(df['日期'])

  15. # 筛选日期大于2022-12-13的行
  16. filtered_df = df[df['日期'] > '2022-12-13']

  18. # 将B列转换为整数格式
  19. filtered_df['B'] = filtered_df['B'].astype(int)

  21. # 找到B列的最大值
  22. max_value = filtered_df['B'].max()

  24. # 找到对应最大值的日期
  25. max_date = filtered_df.loc[filtered_df['B'] == max_value, '日期'].iloc[-1]

  27. print("最大值:", max_value)
  28. print("对应的日期:", max_date.strftime('%Y-%m-%d'))
复制代码


输出结果如下：

  1. 最大值: 12
  2. 对应的日期: 2022-12-19
复制代码

这个结果与您的预期一致：最大值是12，对应的日期是2022-12-19。
小甲鱼最新课程 -> https://ilovefishc.com
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sfqxx
发表于 2023-4-17 17:30:11 | 显示全部楼层
根据用户提供的代码和需求，可以按照以下步骤实现：

1. 修改日期列的数据类型为datetime
2. 以日期为索引，对数据帧进行排序
3. 使用.loc选择大于指定日期范围内的所有行
4. 找到B列最大值及其对应日期

请看下面的代码注释 for more details:


  2. # 导入必要的模块

  4. import pandas as pd


  7. # 定义数据和列名

  9. data = [["004243", "C", "2022-12-22", "", "4"],
  10.         ["162719", "A", "2022-12-20", "8", "6"],
  11.         ["004243", "C", "2022-12-19", "", "12"],
  12.         ["519212", "B", "2022-12-18", "56", "3"],
  13.         ["162411", "E", "2022-12-17", "8", "7"],
  14.         ["007844", "FC", "2022-12-16", "", "8"],
  15.         ["163208", "G", "2022-12-15", "66", "12"],
  16.         ["162719", "A", "2022-12-14", "8", "6"],
  17.         ["519212", "B", "2022-12-12", "56", "3"],
  18.         ["162411", "E", "2022-12-11", "8", "7"],
  19.         ["007844", "FC", "2022-12-10", "", "18"],
  20.         ["163208", "G", "2022-12-8", "66", "12"]]

  22. # 定义列名
  23. columns = ["编号", "名称", "日期", "A", "B"]


  26. # 生成数据帧

  28. df = pd.DataFrame(data, columns=columns)


  31. # 将“日期”列转换为datetime类型
  32. df['日期'] = pd.to_datetime(df['日期'], errors='coerce')


  35. # 根据日期排序，并以日期作为索引
  36. df = df.sort_values(by='日期').set_index('日期')


  39. # 根据指定范围选择行
  40. selected_rows
复制代码

有用请设置最佳答案
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futui
 楼主| 发表于 2023-4-17 17:42:48 | 显示全部楼层
isdkz 发表于 2023-4-17 17:27
为了找到给定时间段内的B列的最大值及对应的日期，我们需要首先筛选出符合条件的行，然后找到最大值以及对 ...

日期序列是这样的，大师你取它的[-1],取到了正确结果，能解释下吗？实在是不理解
2   2022-12-19
6   2022-12-15
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