Pandas: How to find a particular pattern in a dataframe column?

Question

I\'d like to find a particular pattern in a pandas dataframe column, and return the corresponding index values in order to subset the dataframe.

Here\'s a sample dat

Answer 1

The shortest way is finding the index at which the pattern starts. Then you just need to select the three following rows.

In order to find these indexes, a one-liner is enough:

indexes=df[(df.ColA==pattern[0])&(df["ColA"].shift(-1)==pattern[1])&(df["ColA"].shift(-2)==pattern[2])].index

Then do as the other answer says to get the subsets that you want.

Answer 2

for col in df:
    index = df[col][(df[col] == pattern[0]) & (df[col].shift(-1) == pattern[1]) & (df[col].shift(-2) == pattern[2])].index
    if not index.empty: print(index)

Answer 3

Here is a solution:

Check if the pattern was found in any of the columns using rolling. This will give you the last index of the group matching the pattern

matched = df.rolling(len(pattern)).apply(lambda x: all(np.equal(x, pattern)))
matched = matched.sum(axis = 1).astype(bool)   #Sum to perform boolean OR

matched
Out[129]: 
Dates
2017-07-07    False
2017-07-08    False
2017-07-09    False
2017-07-10    False
2017-07-11    False
2017-07-12     True
2017-07-13    False
2017-07-14    False
2017-07-15    False
2017-07-16    False
dtype: bool

For each match, add the indexes of the complete pattern:

idx_matched = np.where(matched)[0]
subset = [range(match-len(pattern)+1, match+1) for match in idx_matched]

Get all the patterns:

result = pd.concat([df.iloc[subs,:] for subs in subset], axis = 0)

result
Out[128]: 
            ColA  ColB
Dates                 
2017-07-10   100    91
2017-07-11    90   107
2017-07-12   105    99

Answer 4

Using the magic of list comprehensions:

[df.index[i - len(pattern)] # Get the datetime index 
 for i in range(len(pattern), len(df)) # For each 3 consequent elements 
 if all(df['ColA'][i-len(pattern):i] == pattern)] # If the pattern matched 

# [Timestamp('2017-07-10 00:00:00')]