我需要设计一个Dataframe的样式:
df = DataFrame({'A':['Bob','Rob','Dob'],'B':['Bob', 'Rob','Dob'],'C':['Bob','Dob','Dob'],'D':['Ben','Ten','Zen'],'E':['Ben','Ten','Zu']}) df A B C D E 0 Bob Bob Bob Ben Ben 1 Rob Rob Dob Ten Ten 2 Dob Dob Dob Zen Zu我需要一次比较A,B,C列,以检查它们是否相等,然后将高亮/颜色应用于不等值。 然后我需要比较列D,E来检查它们是否相等,然后将高亮/颜色应用于不等值
喜欢:
df[['A','B','C']].eq(df.iloc[:, 0], axis=0) A B C 0 True True True 1 True True False 2 True True True我无法将df.style应用于df的子集,然后应用concat。
回复@jezrael回应:
I need to style a Dataframe:
df = DataFrame({'A':['Bob','Rob','Dob'],'B':['Bob', 'Rob','Dob'],'C':['Bob','Dob','Dob'],'D':['Ben','Ten','Zen'],'E':['Ben','Ten','Zu']}) df A B C D E 0 Bob Bob Bob Ben Ben 1 Rob Rob Dob Ten Ten 2 Dob Dob Dob Zen ZuI need to compare columns - A,B, C at once to check if they are equal and then apply a highlight/color to unequal values. Then I need to compare columns D,E to check if they are equal and then apply a highlight/color to unequal values
like:
df[['A','B','C']].eq(df.iloc[:, 0], axis=0) A B C 0 True True True 1 True True False 2 True True TrueI am unable to apply df.style with a subset of df and then concat.
Response to answer by @jezrael:
最满意答案
我相信需要:
def highlight(x): c1 = 'background-color: red' c2 = '' #define groups of columns for compare by first value of group -> #first with A, second with D cols = [['A','B','C'], ['D','E']] #join all masks together m = pd.concat([x[g].eq(x[g[0]], axis=0) for g in cols], axis=1) df1 = pd.DataFrame(c2, index=x.index, columns=x.columns) df1 = df1.where(m, c1) return df1 df.style.apply(highlight, axis=None)
编辑:对于多种颜色是可能的创建字典的颜色与列进行比较:
def highlight(x): c = 'background-color: ' cols = {'red': ['A','B','C'], 'blue':['D','E']} m = pd.concat([x[v].eq(x[v[0]], axis=0).applymap({False:c+k, True:''}.get) for k, v in cols.items()], axis=1) return m
EDIT1:
替代方案:
def highlight(x): c = 'background-color: ' cols = {'red': ['A','B','C'], 'blue':['D','E']} df1 = pd.DataFrame(c, index=x.index, columns=x.columns) for k, v in cols.items(): m = x[v].eq(x[v[0]], axis=0).reindex(columns=x.columns, fill_value=True) df1 = df1.where(m, c+k) return df1 df.style.apply(highlight, axis=None)I believe need:
def highlight(x): c1 = 'background-color: red' c2 = '' #define groups of columns for compare by first value of group -> #first with A, second with D cols = [['A','B','C'], ['D','E']] #join all masks together m = pd.concat([x[g].eq(x[g[0]], axis=0) for g in cols], axis=1) df1 = pd.DataFrame(c2, index=x.index, columns=x.columns) df1 = df1.where(m, c1) return df1 df.style.apply(highlight, axis=None)EDIT: For multiple colors is possible create dictionary by colors with columns for compare:
def highlight(x): c = 'background-color: ' cols = {'red': ['A','B','C'], 'blue':['D','E']} m = pd.concat([x[v].eq(x[v[0]], axis=0).applymap({False:c+k, True:''}.get) for k, v in cols.items()], axis=1) return mEDIT1:
Alternative solution:
def highlight(x): c = 'background-color: ' cols = {'red': ['A','B','C'], 'blue':['D','E']} df1 = pd.DataFrame(c, index=x.index, columns=x.columns) for k, v in cols.items(): m = x[v].eq(x[v[0]], axis=0).reindex(columns=x.columns, fill_value=True) df1 = df1.where(m, c+k) return df1 df.style.apply(highlight, axis=None)更多推荐
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