英文:
Compare row wise elements of a single column. If there are 2 continuous L then select lowest from High column and ignore other. Conversly if 2 L
问题
High D_HIGH D_HIGH_H33 46.57 0 0L1 86.44 68 68H34 56.58 83 83L2 117.91 158 158H36 94.51 186 186L4 123.28 254 254H37 83.20 286 286L
英文:
High D_HIGH D_HIGH_H33 46.57 0 0L0 69.93 42 42H1 86.44 68 68H34 56.58 83 83L35 67.12 125 125L2 117.91 158 158H36 94.51 186 186L3 120.45 245 245H4 123.28 254 254H37 83.20 286 286L
In column D_HIGH_H there is L & H at end.
If there are two continuous H then the one having highest value in High column has to be selected and other has to be ignored(deleted).
If there are two continuous L then the one having lowest value in High column has to be selected and other has to be ignored(deleted).
If the sequence is H,L,H,L then no changes to be made.
Output I want is as follows:
High D_HIGH D_HIGH_H33 46.57 0 0L1 86.44 68 68H34 56.58 83 83L2 117.91 158 158H36 94.51 186 186L4 123.28 254 254H37 83.20 286 286L
I tried various options using list map but did not work out.Also tried with groupby but no logical conclusion.
答案1
得分: 3
以下是代码的翻译部分：
g = ((l := df['D_HIGH_H'].str[-1]) != l.shift()).cumsum()def f(x):if (x['D_HIGH_H'].str[-1] == 'H').any():return x.nlargest(1, 'D_HIGH')return x.nsmallest(1, 'D_HIGH')df.groupby(g, as_index=False).apply(f)
Output:
High D_HIGH D_HIGH_H0 33 46.57 0 0L1 1 86.44 68 68H2 34 56.58 83 83L3 2 117.91 158 158H4 36 94.51 186 186L5 4 123.28 254 254H6 37 83.20 286 286L
英文:
Here's one way:
g = ((l := df['D_HIGH_H'].str[-1]) != l.shift()).cumsum()def f(x):if (x['D_HIGH_H'].str[-1] == 'H').any():return x.nlargest(1, 'D_HIGH')return x.nsmallest(1, 'D_HIGH')df.groupby(g, as_index=False).apply(f)
Output:
High D_HIGH D_HIGH_H0 33 46.57 0 0L1 1 86.44 68 68H2 34 56.58 83 83L3 2 117.91 158 158H4 36 94.51 186 186L5 4 123.28 254 254H6 37 83.20 286 286L
答案2
得分: 2
你可以使用 extract 来获取字母，然后计算一个自定义组，并使用依赖于该字母的函数进行 groupby.apply：
# 提取字母s = df['D_HIGH_H'].str.extract('(\D)$', expand=False)# 按连续字母分组# 根据字母类型获取 idxmin/idxmaxkeep = (df['High'].groupby([s, s.ne(s.shift()).cumsum()], sort=False).apply(lambda x: x.idxmin() if x.name[0] == 'L' else x.idxmax()).tolist())out = df.loc[keep]
输出：
High D_HIGH D_HIGH_H33 46.57 0 0L1 86.44 68 68H34 56.58 83 83L2 117.91 158 158H36 94.51 186 186L4 123.28 254 254H37 83.20 286 286L
英文:
You can use extract to get the letter, then compute a custom group and groupby.apply with a function that depends on the letter:
# extract letters = df['D_HIGH_H'].str.extract('(\D)$', expand=False)# group by successive letters# get the idxmin/idxmax depending on the type of letterkeep = (df['High'].groupby([s, s.ne(s.shift()).cumsum()], sort=False).apply(lambda x: x.idxmin() if x.name[0] == 'L' else x.idxmax()).tolist())out = df.loc[keep]
Output:
High D_HIGH D_HIGH_H33 46.57 0 0L1 86.44 68 68H34 56.58 83 83L2 117.91 158 158H36 94.51 186 186L4 123.28 254 254H37 83.20 286 286L
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