將值轉換爲％

Question

我有一個數據框，下面有來自每家商店的肉，蔬菜和麪包的銷售額。我想將這些值轉換爲％，例如，Store N的值將變爲74％，7％和19％。換句話說，就商店N的總銷售額而言，74％是肉類銷售額的百分比。什麼是最簡單的方法呢？

import pandas as pd 

df=pd.DataFrame({'Store':['N','S','E','W'] 
        ,'Meat':[200,250,100,400] 
        ,'Veg':[20,100,30,80] 
        ,'Bread':[50,230,150,100]}) 
df=df[['Store','Meat','Veg','Bread']]  

Answer 1

您也可以用pandas.apply lambda函數使用：

df.ix[:, 1:]=df.ix[:,1:].apply(lambda x: x*100/x.sum(), axis=1) 

它給你：

Store  Meat  Veg  Bread 
0  N 74.074074 7.407407 18.518519 
1  S 43.103448 17.241379 39.655172 
2  E 35.714286 10.714286 53.571429 
3  W 68.965517 13.793103 17.241379 

Answer 2

可以只是手動計算百分比：

df['MeatPerc'] = df['Meat']/df['Meat'].sum()

Answer 3

在不使用循環A純大熊貓的解決辦法是：

df.ix[:, 1:] = (df.ix[:, 1:].T/df.ix[:, 1:].sum(1)).T 
print(df) 

結果：

Store  Meat  Veg  Bread 
0  N 0.740741 0.074074 0.185185 
1  S 0.431034 0.172414 0.396552 
2  E 0.357143 0.107143 0.535714 
3  W 0.689655 0.137931 0.172414 

Answer 4

你可以先用set_index列Store，然後通過sumdiv分最後reset_index：

df.set_index('Store', inplace=True) 
df = df.div(df.sum(1), axis=0) 
print (df.reset_index()) 
    Store  Meat  Veg  Bread 
0  N 0.740741 0.074074 0.185185 
1  S 0.431034 0.172414 0.396552 
2  E 0.357143 0.107143 0.535714 
3  W 0.689655 0.137931 0.172414 

通過ix或iloc與選擇另一種解決方案：

df.ix[:,'Meat':] = df.ix[:,'Meat':].div(df.ix[:,'Meat':].sum(1), axis=0) 
print (df) 
    Store  Meat  Veg  Bread 
0  N 0.740741 0.074074 0.185185 
1  S 0.431034 0.172414 0.396552 
2  E 0.357143 0.107143 0.535714 
3  W 0.689655 0.137931 0.172414 

---

df.iloc[:,1:] = df.iloc[:,1:].div(df.iloc[:,1:].sum(1), axis=0) 
print (df) 
    Store  Meat  Veg  Bread 
0  N 0.740741 0.074074 0.185185 
1  S 0.431034 0.172414 0.396552 
2  E 0.357143 0.107143 0.535714 
3  W 0.689655 0.137931 0.172414 

時序：

In [187]: %timeit (jez1(df)) 
100 loops, best of 3: 4.07 ms per loop 

In [188]: %timeit (jez2(df1)) 
100 loops, best of 3: 5.61 ms per loop 

In [189]: %timeit (jez3(df2)) 
100 loops, best of 3: 5.44 ms per loop 

In [190]: %timeit (ric(df3)) 
100 loops, best of 3: 6.18 ms per loop 

In [191]: %timeit (ogi(df4)) 
1 loop, best of 3: 2.2 s per loop 

代碼定時 S：

#random dataframe 
np.random.seed(100) 

#10 data columns + first Store col, 10k rows 
df = pd.DataFrame(np.random.randint(10, size=(10000,10)), columns=list('ABCDEFGHIJ')) 
df.index = 'a' + df.index.astype(str) 
df = df.reset_index().rename(columns={'index':'Store'}) 
print (df) 
df1, df2, df3, df4 = df.copy(), df.copy(), df.copy(), df.copy() 

def jez1(df): 
    df = df.set_index('Store') 
    df = 100 * df.div(df.sum(1), axis=0) 
    return (df.reset_index()) 


def jez2(df): 
    df.ix[:,'A':] = df.ix[:,'A':].div(df.ix[:,'A':].sum(1), axis=0) 
    return df 
def jez3(df):  
    df.iloc[:,1:] = df.iloc[:,1:].div(df.iloc[:,1:].sum(1), axis=0) 
    return df 

def ric(df):  
    df.ix[:, 1:] = (df.ix[:, 1:].T/df.ix[:, 1:].sum(1)).T 
    return df 

def ogi(df):  
    df.ix[:, 1:]=df.ix[:,1:].apply(lambda x: x/x.sum(), axis=1) 
    return df  

print (jez1(df)) 
print (jez2(df1)) 
print (jez3(df2)) 
print (ric(df3)) 
print (ogi(df4))