有一种方法可以找到在切换到另一个索引值之前的每个最大值吗？

huangapple
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英文:

Is there a method to find each maximum value before changing to another index value?

问题

以下是您要翻译的内容：

  1. for i, row in df.iloc[:-1].iterrows():
  2.    if df['index_a'][i] == 0:
  3.      df['all_max'][i] = df['column_b'][i].max()
  4.    else:
  5.      df['all_max'][i] = df['column_b'][i].max()

注意：您提供的代码部分不需要翻译，因此我只提供了代码的翻译部分。

英文:

I want to get the max for each index before its changing to another index. As shown in below dataframe example

index_a column_b all_max
0 10 -
0 20 -
0 30 30
1 50 50
1 30 -
1 20 -
1 10 -
0 70 70
0 60 -
0 40 -

... (so on)

but instead i get results like this using the function i mentioned below

index_a column_b all_max
0 10 70
0 20 70
0 30 70
1 50 70
1 30 70
1 20 70
1 10 70
0 70 70
0 60 70
0 40 70

... (so on)

the index row is not fixed repetition, some have more 1s or 0s.

I have tried using the .max() function but it only provide me for the max value inside the Column B

  1. for i, row in df.iloc[:-1].iterrows():
  2.    if df['index_a'][i] == 0:
  3.      df['all_max'][i] = df['column_b'][i].max()
  4.    else:
  5.      df['all_max'][i] = df['column_b'][i].max()

答案1

得分: 1

使用groupby.transform在连续的分组上获取每个组的最大值作为广播系列，然后使用where来识别最大行并分配值，否则为-：

  1. group = df['index_a'].ne(df['index_a'].shift()).cumsum()
  2. m = df.groupby(group)['column_b'].transform('max').eq(df['column_b'])
  4. df['all_max'] = df['column_b'].where(m, '-')

输出结果：

  1.    index_a  column_b all_max
  2. 0        0        10       -
  3. 1        0        20       -
  4. 2        0        30      30
  5. 3        1        50      50
  6. 4        1        30       -
  7. 5        1        20       -
  8. 6        1        10       -
  9. 7        0        70      70
  10. 8        0        60       -
  11. 9        0        40       -
英文:

Use groupby.tranform on successive groups to get the max per group as a broadcasted Series, then identify the max rows and assign the value with where, else a -:

  1. group = df['index_a'].ne(df['index_a'].shift()).cumsum()
  2. m = df.groupby(group)['column_b'].transform('max').eq(df['column_b'])
  4. df['all_max'] = df['column_b'].where(m, '-')

Output:

  1.    index_a  column_b all_max
  2. 0        0        10       -
  3. 1        0        20       -
  4. 2        0        30      30
  5. 3        1        50      50
  6. 4        1        30       -
  7. 5        1        20       -
  8. 6        1        10       -
  9. 7        0        70      70
  10. 8        0        60       -
  11. 9        0        40       -

答案2

得分: 0

如果需要获取每个连续分组的所有最大值，请使用GroupBy.transform来比较移位值与累积和，然后使用Series.where根据原始的column_b进行比较，如果不匹配，则赋值为-：

  1. m = (df.groupby(df['index_a'].ne(df['index_a'].shift()).cumsum())['column_b']
  2.        .transform('max')
  3.        .eq(df['column_b']))
  5. df['all_max'] = df['column_b'].where(m, '-')

但如果只需要匹配每个相同分组中的第一个最大值，请使用：

  1. i = df.groupby(df['index_a'].ne(df['index_a'].shift()).cumsum())['column_b'].idxmax()
  3. df['all_max'] = '-'
  4. df.loc[i, 'all_max'] = df['column_b']

你可以看到在更改的数据中的差异：

  1. print(df)
  2.    index_a  column_b
  3. 0        0        10
  4. 1        0        30 <- 第一个0组中有2个最大值
  5. 2        0        30
  6. 3        1        50
  7. 4        1        30
  8. 5        1        20
  9. 6        1        10
  10. 7        0        70
  11. 8        0        60
  12. 9        0        40

以下是两种方法的应用：

  1. m = (df.groupby(df['index_a'].ne(df['index_a'].shift()).cumsum())['column_b']
  2.        .transform('max')
  3.        .eq(df['column_b']))
  5. df['all_max1'] = df['column_b'].where(m, '-')
  7. i = df.groupby(df['index_a'].ne(df['index_a'].shift()).cumsum())['column_b'].idxmax()
  9. df['all_max2'] = '-'
  10. df.loc[i, 'all_max2'] = df['column_b']

最终的输出如下：

  1. print(df)
  2.    index_a  column_b all_max1 all_max2
  3. 0        0        10        -        -
  4. 1        0        30       30       30
  5. 2        0        30       30        -
  6. 3        1        50       50       50
  7. 4        1        30        -        -
  8. 5        1        20        -        -
  9. 6        1        10        -        -
  10. 7        0        70       70       70
  11. 8        0        60        -        -
英文:

If need all maximum values per consecutive groups use GroupBy.transform by groups by compare shifted values with cumulative sums, compare by original column_b and assign - if no match in Series.where:

  1. m = (df.groupby(df[&#39;index_a&#39;].ne(df[&#39;index_a&#39;].shift()).cumsum())[&#39;column_b&#39;]
  2.        .transform(&#39;max&#39;)
  3.        .eq(df[&#39;column_b&#39;]))
  4.  
  5. df[&#39;all_max&#39;] = df[&#39;column_b&#39;].where(m, &#39;-&#39;)
  7. print (df)
  8.    index_a  column_b all_max
  9. 0        0        10       -
  10. 1        0        20       -
  11. 2        0        30      30
  12. 3        1        50      50
  13. 4        1        30       -
  14. 5        1        20       -
  15. 6        1        10       -
  16. 7        0        70      70
  17. 8        0        60       -
  18. 9        0        40       -

But if need match only first maximal value per same groups use:

  1. i = df.groupby(df[&#39;index_a&#39;].ne(df[&#39;index_a&#39;].shift()).cumsum())[&#39;column_b&#39;].idxmax()
  3. df[&#39;all_max&#39;] = &#39;-&#39;
  4. df.loc[i, &#39;all_max&#39;] = df[&#39;column_b&#39;]
  6. print (df)
  7.    index_a  column_b all_max
  8. 0        0        10       -
  9. 1        0        20       -
  10. 2        0        30      30
  11. 3        1        50      50
  12. 4        1        30       -
  13. 5        1        20       -
  14. 6        1        10       -
  15. 7        0        70      70
  16. 8        0        60       -
  17. 9        0        40       -

You can see difference in changed data:

  1. print (df)
  2.    index_a  column_b
  3. 0        0        10
  4. 1        0        30 &lt;- 2 maximums per first 0 group
  5. 2        0        30
  6. 3        1        50
  7. 4        1        30
  8. 5        1        20
  9. 6        1        10
  10. 7        0        70
  11. 8        0        60
  12. 9        0        40
  14. m = (df.groupby(df[&#39;index_a&#39;].ne(df[&#39;index_a&#39;].shift()).cumsum())[&#39;column_b&#39;]
  15.        .transform(&#39;max&#39;)
  16.        .eq(df[&#39;column_b&#39;]))
  17.  
  18. df[&#39;all_max1&#39;] = df[&#39;column_b&#39;].where(m, &#39;-&#39;)
  21. i = df.groupby(df[&#39;index_a&#39;].ne(df[&#39;index_a&#39;].shift()).cumsum())[&#39;column_b&#39;].idxmax()
  23. df[&#39;all_max2&#39;] = &#39;-&#39;
  24. df.loc[i, &#39;all_max2&#39;] = df[&#39;column_b&#39;]
  1. print (df)
  2.    index_a  column_b all_max1 all_max2
  3. 0        0        10        -        -
  4. 1        0        30       30       30
  5. 2        0        30       30        -
  6. 3        1        50       50       50
  7. 4        1        30        -        -
  8. 5        1        20        -        -
  9. 6        1        10        -        -
  10. 7        0        70       70       70
  11. 8        0        60        -        -

答案3

得分: 0

创建虚拟组以区分index_a列中相同的标识符：

  1. df['all_max'] = (df.groupby(df['index_a'].ne(df['index_a'].shift()).cumsum())['column_b']
  2.                    .transform('max'))
  3. print(df)
  5. # 输出
  6.    index_a  column_b  all_max
  7. 0        0        10       30
  8. 1        0        20       30
  9. 2        0        30       30
  10. 3        1        50       50
  11. 4        1        30       50
  12. 5        1        20       50
  13. 6        1        10       50
  14. 7        0        70       70
  15. 8        0        60       70
  16. 9        0        40       70
英文:

Create virtual groups to distinguish same identifier in index_a column:

  1. df[&#39;all_max&#39;] = (df.groupby(df[&#39;index_a&#39;].ne(df[&#39;index_a&#39;].shift()).cumsum())[&#39;column_b&#39;]
  2.                    .transform(&#39;max&#39;))
  3. print(df)
  5. # Output
  6.    index_a  column_b  all_max
  7. 0        0        10       30
  8. 1        0        20       30
  9. 2        0        30       30
  10. 3        1        50       50
  11. 4        1        30       50
  12. 5        1        20       50
  13. 6        1        10       50
  14. 7        0        70       70
  15. 8        0        60       70
  16. 9        0        40       70

huangapple
  • 本文由 发表于 2023年2月8日 18:43:23
  • 转载请务必保留本文链接：https://go.coder-hub.com/75384622.html
  • dataframe
  • pandas
  • python
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