英文:
How to check the row trend and add the difference and difference percentage in separate columns for the failing cases
问题
以下是翻译好的代码部分：
import pandas as pdimport numpy as npd = {'Cell': ['A', 'B', 'C', 'D', 'E'], 'D1': [5, 2, 2, 6, 6], 'D2': [np.nan, 5, 6, np.nan, 3], 'D3': [7, np.nan, 5, 5, np.nan], 'D6': [17, 3, np.nan, np.nan, 2]}df = pd.DataFrame(d)
希望对你有所帮助。如需进一步的翻译或解释，请随时提出。
英文:
An extension to the problem statement
https://stackoverflow.com/questions/75412399/how-to-check-each-row-trend-with-some-tolerance-by-ignoring-the-np-nan-values-in
import pandas as pdimport numpy as npd = {'Cell':['A','B','C','D','E'],'D1':[5, 2, 2, 6,6], 'D2':[np.nan, 5, 6, np.nan,3], 'D3':[7,np.nan, 5, 5,np.nan], 'D6':[17, 3, np.nan,np.nan,2]}df = pd.DataFrame(d)Cell D1 D2 D3 D60 A 5 NaN 7.0 17.01 B 2 5.0 NaN 3.02 C 2 6.0 5.0 NaN3 D 6 NaN 5.0 NaN4 E 6 3.0 NaN 2.0
i want output like this with additional columns diff and diff% along with is_increasing and failing columns
Cell D1 D2 D3 D6 is_increasing? failing diff diff%0 A 5 NaN 7.0 17.0 True NaN NaN NaN1 B 2 5.0 NaN 3.0 False [D6] [-2] [40%]2 C 2 6.0 5.0 NaN False [D3] [-1] [16.6%]3 D 6 NaN 5.0 NaN False [D3] [-1] [16.6%]4 E 6 3.0 NaN 2.0 False [D2, D6] [-3,-1] [50%,33%]
Explanation of the columns:
is_increasing --> whether the values are strictly increasing or notfailing --> columns whether strictly increasing is not followed when compared with previous valuediff --> difference of the values where there is failing casesdiff% --> difference in terms of percentages for the failing cases
between (6,5) numbers in the columns
diff column --> 5-6=-1diff%--> 1-(5/6)=16.6%
Please let me the solution to this problem, i tried different ways but not able to come up with solution.
答案1
得分: 1
以下是您要翻译的内容：
# 仅筛选相关列# 使用任何方法df2 = df.drop(columns='Cell')d = (df2.ffill(axis=1).diff(axis=1))m = (d.where(df2.notna()).lt(0))df['is_increasing'] = ~m.any(axis=1)df['failing'] = (m.mul(df2.columns).where(m).stack().groupby(level=0).agg(list))df['diff'] = (d.where(m).stack().groupby(level=0).agg(list))p = (df2.ffill(axis=1).pct_change(axis=1).mul(-100).round(2))df['diff%'] = (p.where(m).stack().groupby(level=0).agg(list))print(df)
输出：
Cell D1 D2 D3 D6 is_increasing failing diff diff%0 A 5 NaN 7.0 17.0 True NaN NaN NaN1 B 2 5.0 NaN 3.0 False [D6] [-2.0] [40.0]2 C 2 6.0 5.0 NaN False [D3] [-1.0] [16.67]3 D 6 NaN 5.0 NaN False [D3] [-1.0] [16.67]4 E 6 3.0 NaN 2.0 False [D2, D6] [-3.0, -1.0] [50.0, 33.33]
请注意，代码中的注释部分未被翻译。
英文:
You can use:
# filter only relevant columns# use any methoddf2 = df.drop(columns='Cell')d = (df2.ffill(axis=1).diff(axis=1))m = (d.where(df2.notna()).lt(0))df['is_increasing'] = ~m.any(axis=1)df['failing'] = (m.mul(df2.columns).where(m).stack().groupby(level=0).agg(list))df['diff'] = (d.where(m).stack().groupby(level=0).agg(list))p = (df2.ffill(axis=1).pct_change(axis=1).mul(-100).round(2))df['diff%'] = (p.where(m).stack().groupby(level=0).agg(list))print(df)
Output:
Cell D1 D2 D3 D6 is_increasing failing diff diff%0 A 5 NaN 7.0 17.0 True NaN NaN NaN1 B 2 5.0 NaN 3.0 False [D6] [-2.0] [40.0]2 C 2 6.0 5.0 NaN False [D3] [-1.0] [16.67]3 D 6 NaN 5.0 NaN False [D3] [-1.0] [16.67]4 E 6 3.0 NaN 2.0 False [D2, D6] [-3.0, -1.0] [50.0, 33.33]
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