Comparing previous row values in Pandas DataFrame

1) pandas approach: Use diff:

df['match'] = df['col1'].diff().eq(0)

---

2) numpy approach: Use np.ediff1d.

df['match'] = np.ediff1d(df['col1'].values, to_begin=np.NaN) == 0

Both produce:

Timings: (for the same DF used by @jezrael)

%timeit df.col1.eq(df.col1.shift())
1000 loops, best of 3: 731 µs per loop

%timeit df['col1'].diff().eq(0)
1000 loops, best of 3: 405 µs per loop

Here's a NumPy arrays based approach using slicing that lets us use the views into the input array for efficiency purposes -

def comp_prev(a):
    return np.concatenate(([False],a[1:] == a[:-1]))

df['match'] = comp_prev(df.col1.values)

Sample run -

In [48]: df['match'] = comp_prev(df.col1.values)

In [49]: df
Out[49]: 
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

Runtime test -

In [56]: data={'col1':[1,3,3,1,2,3,2,2]}
    ...: df0=pd.DataFrame(data,columns=['col1'])
    ...: 

#@jezrael's soln1
In [57]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [58]: %timeit df['match'] = df.col1 == df.col1.shift() 
1000 loops, best of 3: 1.53 ms per loop

#@jezrael's soln2
In [59]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [60]: %timeit df['match'] = df.col1.eq(df.col1.shift())
1000 loops, best of 3: 1.49 ms per loop

#@Nickil Maveli's soln1   
In [61]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [64]: %timeit df['match'] = df['col1'].diff().eq(0) 
1000 loops, best of 3: 1.02 ms per loop

#@Nickil Maveli's soln2
In [65]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [66]: %timeit df['match'] = np.ediff1d(df['col1'].values, to_begin=np.NaN) == 0
1000 loops, best of 3: 1.52 ms per loop

# Posted approach in this post
In [67]: df = pd.concat([df0]*10000).reset_index(drop=True)

In [68]: %timeit df['match'] = comp_prev(df.col1.values)
1000 loops, best of 3: 376 µs per loop

You need eq with shift:

df['match'] = df.col1.eq(df.col1.shift())
print (df)
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

Or instead eq use ==, but it is a bit slowier in large DataFrame:

df['match'] = df.col1 == df.col1.shift()
print (df)
   col1  match
0     1  False
1     3  False
2     3   True
3     1  False
4     2  False
5     3  False
6     2  False
7     2   True

Timings:

import pandas as pd
data={'col1':[1,3,3,1,2,3,2,2]}
df=pd.DataFrame(data,columns=['col1'])
print (df)
#[80000 rows x 1 columns]
df = pd.concat([df]*10000).reset_index(drop=True)

df['match'] = df.col1 == df.col1.shift()
df['match1'] = df.col1.eq(df.col1.shift())
print (df)

In [208]: %timeit df.col1.eq(df.col1.shift())
The slowest run took 4.83 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 933 µs per loop

In [209]: %timeit df.col1 == df.col1.shift()
1000 loops, best of 3: 1 ms per loop