熊貓隨機取代k百分號

Question

具有簡單的熊貓數據幀和2列例如id和value其中value是0或1我想隨機用0代替所有value==1的10%。

我該如何用大熊貓實現這種行爲？

Answer 1

pandas答案

• 使用query得到過濾df只有value == 1
• 使用sample(frac=.1)採取的那些
• 10％使用結果的指標分配零

---

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

---

替代numpy答案

• 得到的其中df['value']是1
• 分配10點％的零和90周％的人

的隨機陣列布爾數組

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

Answer 2

你也許可以使用numpy.random.choice：

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

Answer 3

這裏有一個np.random.choice方法NumPy的 -

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 

採樣運行 -

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 

---

另外，更多的大熊貓的方法 -

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

---

運行測試

所有的方法發佈迄今爲止 -

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 

計時 -

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