Question

我有一个Pandas DataFrame，其列为具有2个级别的MultiIndex，如下所示：

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我只是想将“旅行”和“食物”列中每个列的差异存储到新的顶级列中-例如'diff'-在“昨天”和“旅行”旁边

index = ['monday','tuesday','wednesday']
tuples = [('yesterday','travel'),('yesterday','food'),('today','travel'),('today','food')]
columns = pd.MultiIndex.from_tuples(tuples,names=[None,'category'])
df = pd.DataFrame(np.random.randint(low=0, high=10, size=(3, 4)), index=index, columns=columns)

将返回我感兴趣的基础DataFrame，但我无法弄清楚如何将其正确放置在整个DataFrame中

类似：

diff = t['today'] - t['yesterday']

产生有趣（但不正确）的结果

Answer 1

一种方法可能是将diff的列设置为MultiIndex，例如：

diff = df['today'] - df['yesterday']
diff.columns = pd.MultiIndex.from_tuples([('diff',col) for col in diff.columns])

，然后当您使用concat时，它会给出：

print (pd.concat([df,diff],axis=1))
          yesterday       today        diff     
category     travel food travel food travel food
monday            8    7      7    1     -1   -6
tuesday           1    3      0    8     -1    5
wednesday         6    4      5    6     -1    2

编辑：不使用MultiIndex的另一种方法可能是直接执行创建列的操作：

df[[('diff','travel'),('diff','food')]] = df['today'] - df['yesterday']

还有一种更通用的方法，您可以使用get_level_values

df[[('diff',col) for col in df.columns.get_level_values(1).unique()]] = df['today'] - df['yesterday']

多索引的Pandas Dataframe中2列之间的差异

1 个答案: