Question

具有2列的简单pandas数据帧，例如id和value其中value为0或1我希望随机替换所有10%中的value==1 { {1}}。

如何使用pandas实现此行为？

Answer 1

pandas回答

使用query仅使用df

value == 1

使用sample(frac=.1)获取10％的
使用结果的索引指定零

df.loc[
    df.query('value == 1').sample(frac=.1).index,
    'value'
] = 0

替代numpy回答

获取df['value']为1
指定10％零和90％

v = df.value.values == 1
df.loc[v, 'value'] = np.random.choice((0, 1), v.sum(), p=(.1, .9))

Answer 2

这是使用np.random.choice -

的NumPy方法

a = df.value.values  # get a view into value col
idx = np.flatnonzero(a) # get the nonzero indices

# Finally select unique 10% from those indices and set 0s there
a[np.random.choice(idx,size=int(0.1*len(idx)),replace=0)] = 0

示例运行 -

In [237]: df = pd.DataFrame(np.random.randint(0,2,(100,2)),columns=['id','value'])

In [238]: (df.value==1).sum() # Original Count of 1s in df.value column
Out[238]: 53

In [239]: a = df.value.values

In [240]: idx = np.flatnonzero(a)

In [241]: a[np.random.choice(idx,size=int(0.1*len(idx)),replace=0)] = 0

In [242]: (df.value==1).sum() # New count of 1s in df.value column
Out[242]: 48

或者，更多的熊猫方法 -

idx = np.flatnonzero(df['value'])
df.ix[np.random.choice(idx,size=int(0.1*len(idx)),replace=0),'value'] = 0

运行时测试

到目前为止发布的所有方法 -

def f1(df):  #@piRSquared's soln1
    df.loc[df.query('value == 1').sample(frac=.1).index,'value'] = 0

def f2(df):  #@piRSquared's soln2
    v = df.value.values == 1
    df.loc[v, 'value'] = np.random.choice((0, 1), v.sum(), p=(.1, .9))

def f3(df): #@Roman Pekar's soln
    idx = df.index[df.value==1]
    df.loc[np.random.choice(idx, size=idx.size/10, replace=False)].value = 0

def f4(df): #@Mine soln1
    a = df.value.values
    idx = np.flatnonzero(a)
    a[np.random.choice(idx,size=int(0.1*len(idx)),replace=0)] = 0

def f5(df): #@Mine soln2
    idx = np.flatnonzero(df['value'])
    df.ix[np.random.choice(idx,size=int(0.1*len(idx)),replace=0),'value'] = 0

计时 -

In [2]: # Setup inputs
   ...: df = pd.DataFrame(np.random.randint(0,2,(10000,2)),columns=['id','value'])
   ...: df1 = df.copy()
   ...: df2 = df.copy()
   ...: df3 = df.copy()
   ...: df4 = df.copy()
   ...: df5 = df.copy()
   ...: 

In [3]: # Timings
   ...: %timeit f1(df1)
   ...: %timeit f2(df2)
   ...: %timeit f3(df3)
   ...: %timeit f4(df4)
   ...: %timeit f5(df5)
   ...: 
100 loops, best of 3: 3.96 ms per loop
1000 loops, best of 3: 844 µs per loop
1000 loops, best of 3: 1.62 ms per loop
10000 loops, best of 3: 163 µs per loop
1000 loops, best of 3: 663 µs per loop

Answer 3

您可以使用numpy.random.choice：

>>> idx = df.index[df.value==1]
>>> df.loc[np.random.choice(idx, size=idx.size/10, replace=False)].value = 0

大熊猫随机取代k％

3 个答案: