比较 table 的列并将结果放入另一个 table
Compare the column of table and put result in another table
我有一个如下所示的数据框:
Date Input Val1 Val2
1-Dec X 10 5
2-Dec Y 15
3-Dec Z 4 5
4-Dec A 10
我希望以这种方式输出,如果 Val1 存在,则在输出中显示 val1,否则将在输出中显示 Val2,所以我的输出将是:
Date Input Val1 Val2 Output
1-Dec X 10 5 10
2-Dec Y 15 15 since Val1 is missing so took Val2 in output
3-Dec Z 4 5 4
4-Dec A 10 10 since Val1 is missing so took Val2 in output
我 tred pd.Concat 但它没有给出正确的输出
最简单的就是使用Series.fillna
:
df['Output'] = df['Val1'].fillna(df['Val2'])
测试缺失值的解决方案是numpy.where
:
df['Output'] = np.where(df['Val1'].isna(), df['Val2'], df['Val1'])
我有一个如下所示的数据框:
Date Input Val1 Val2
1-Dec X 10 5
2-Dec Y 15
3-Dec Z 4 5
4-Dec A 10
我希望以这种方式输出,如果 Val1 存在,则在输出中显示 val1,否则将在输出中显示 Val2,所以我的输出将是:
Date Input Val1 Val2 Output
1-Dec X 10 5 10
2-Dec Y 15 15 since Val1 is missing so took Val2 in output
3-Dec Z 4 5 4
4-Dec A 10 10 since Val1 is missing so took Val2 in output
我 tred pd.Concat 但它没有给出正确的输出
最简单的就是使用Series.fillna
:
df['Output'] = df['Val1'].fillna(df['Val2'])
测试缺失值的解决方案是numpy.where
:
df['Output'] = np.where(df['Val1'].isna(), df['Val2'], df['Val1'])