将值设置为DataFrame切片的副本-SettingvaluetoacopyofasliceofaDataFrame

8  

One approach with np.where -

使用np.where的一种方法 -

mask = df.Category.values=='small'
df['Discount'] = np.where(mask,df.price*0.01, df.price*0.02)

Another way to put things a bit differently -

让事情有点不同的另一种方式 -

df['Discount'] = df.price*0.01
df['Discount'][df.Category.values!='small'] *= 2

For performance, you might want to work with array data, so we could use df.price.values instead at places where df.price was used.

为了提高性能,您可能希望使用数组数据,因此我们可以在使用df.price的地方使用df.price.values。

Benchmarking

Approaches -

def app1(df): # Proposed app#1 here
    mask = df.Category.values=='small'
    df_price = df.price.values
    df['Discount'] = np.where(mask,df_price*0.01, df_price*0.02)
    return df

def app2(df): # Proposed app#2 here
    df['Discount'] = df.price.values*0.01
    df['Discount'][df.Category.values!='small'] *= 2
    return df

def app3(df): # @piRSquared's soln
    df.assign(
    Discount=((1 - (df.Category.values == 'small')) + 1) / 100 * df.price.values)
    return df

def app4(df): # @MaxU's soln
    df.assign(Discount=df.price * df.Category.map({'small':0.01}).fillna(0.02))
    return df

Timings -

1) Large dataset :

1)大数据集:

In [122]: df
Out[122]: 
  Category  ID  price  Discount
0    small   1     25      0.25
1   medium   2     30      0.60
2   medium   3     34      0.68
3    small   4     40      0.40

In [123]: df1 = pd.concat([df]*1000,axis=0)
     ...: df2 = pd.concat([df]*1000,axis=0)
     ...: df3 = pd.concat([df]*1000,axis=0)
     ...: df4 = pd.concat([df]*1000,axis=0)
     ...: 

In [124]: %timeit app1(df1)
     ...: %timeit app2(df2)
     ...: %timeit app3(df3)
     ...: %timeit app4(df4)
     ...: 
1000 loops, best of 3: 209 µs per loop
10 loops, best of 3: 63.2 ms per loop
1000 loops, best of 3: 351 µs per loop
1000 loops, best of 3: 720 µs per loop

2) Very large dataset :

2)非常大的数据集:

In [125]: df1 = pd.concat([df]*10000,axis=0)
     ...: df2 = pd.concat([df]*10000,axis=0)
     ...: df3 = pd.concat([df]*10000,axis=0)
     ...: df4 = pd.concat([df]*10000,axis=0)
     ...: 

In [126]: %timeit app1(df1)
     ...: %timeit app2(df2)
     ...: %timeit app3(df3)
     ...: %timeit app4(df4)
     ...: 
1000 loops, best of 3: 758 µs per loop
1 loops, best of 3: 2.78 s per loop
1000 loops, best of 3: 1.37 ms per loop
100 loops, best of 3: 2.57 ms per loop

Further boost with data reuse -

数据重用进一步提升 -

def app1_modified(df):
    mask = df.Category.values=='small'
    df_price = df.price.values*0.01
    df['Discount'] = np.where(mask,df_price, df_price*2)
    return df

Timings -

In [133]: df1 = pd.concat([df]*10000,axis=0)
     ...: df2 = pd.concat([df]*10000,axis=0)
     ...: df3 = pd.concat([df]*10000,axis=0)
     ...: df4 = pd.concat([df]*10000,axis=0)
     ...: 

In [134]: %timeit app1(df1)
1000 loops, best of 3: 699 µs per loop

In [135]: %timeit app1_modified(df1)
1000 loops, best of 3: 655 µs per loop

4  

Also using some numpy

也使用一些numpy

df.assign(
    Discount=((1 - (df.Category.values == 'small')) + 1) / 100 * df.price.values)

  Category  ID  price  Discount
0    small   1     25      0.25
1   medium   2     30      0.60
2   medium   3     34      0.68
3    small   4     40      0.40

The operative component being

操作组件是

(1 - (df.Category.values == 'small')) + 1) / 100 * df.price.values

This generates a single boolean array and performs simple arithmetic on it in order to get .01 and .02.

这会生成一个布尔数组,并对其执行简单的算术,以获得.01和.02。

naive time testing over given data

对给定数据进行天真时间测试

Thx @Divakar for pointing this out For those using python 2.x, you'll need to force the float issue by using this.

Thx @Divakar指出这一点对于那些使用python 2.x的人来说,你需要通过使用它来强制浮动问题。

df.assign(
    Discount=((1 - (df.Category.values == 'small')) + 1) / 100. * df.price.values)

4  

Here is yet another Pandas approach:

这是另一个熊猫方法:

In [67]: df.assign(Discount=df.price * df.Category.map({'small':0.01}).fillna(0.02))
Out[67]:
  Category  ID  price  Discount
0    small   1     25      0.25
1   medium   2     30      0.60
2   medium   3     34      0.68
3    small   4     40      0.40