Pandas Randomly Replace K Percent

having a simple pandas data frame with 2 columns e.g. id and value where value is either 0 or 1 I would like to randomly replace 10% of all value==1 with 0. How can I achieve this

Solution 1:

pandas answer

• use query to get filtered df with only value == 1
• use sample(frac=.1) to take 10% of those
• use the index of the result to assign zero

---

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

---

alternative numpy answer

• get boolean array of where df['value'] is 1
• assign random array of 10% zeros and 90% ones

---

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

Solution 2:

Here's a NumPy approach with np.random.choice -

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

Sample run -

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

---

Alternatively, a bit more pandas approach -

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

---

Runtime test

All approaches posted thus far -

deff1(df):  #@piRSquared's soln1
    df.loc[df.query('value == 1').sample(frac=.1).index,'value'] = 0deff2(df):  #@piRSquared's soln2
    v = df.value.values == 1
    df.loc[v, 'value'] = np.random.choice((0, 1), v.sum(), p=(.1, .9))

deff3(df): #@Roman Pekar's soln
    idx = df.index[df.value==1]
    df.loc[np.random.choice(idx, size=idx.size/10, replace=False)].value = 0deff4(df): #@Mine soln1
    a = df.value.values
    idx = np.flatnonzero(a)
    a[np.random.choice(idx,size=int(0.1*len(idx)),replace=0)] = 0deff5(df): #@Mine soln2
    idx = np.flatnonzero(df['value'])
    df.ix[np.random.choice(idx,size=int(0.1*len(idx)),replace=0),'value'] = 0

Timings -

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

Solution 3:

you can probably use numpy.random.choice:

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