Pandas Dataframe check if a value exists using regex

There are two solutions:

1. df.col.apply method is more straightforward but also a little bit slower:

   In [1]: import pandas as pd
   
   In [2]: import re
   
   In [3]: df = pd.DataFrame({'col1':[1,2,3,4,5], 'col2':['admin', 'aa', 'bb', 'c_admin_d', 'ee_admin']})
   
   In [4]: df
   Out[4]: 
      col1       col2
   0     1      admin
   1     2         aa
   2     3         bb
   3     4  c_admin_d
   4     5   ee_admin
   
   In [5]: r = re.compile(r'.*(admin).*')
   
   In [6]: df.col2.apply(lambda x: bool(r.match(x)))
   Out[6]: 
   0     True
   1    False
   2    False
   3     True
   4     True
   Name: col2, dtype: bool
   
   In [7]: %timeit -n 100000 df.col2.apply(lambda x: bool(r.match(x)))
   167 µs ± 1.02 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
   

2. np.vectorize method require import numpy, but it's more efficient (about 4 times faster in my timeit test).

   In [1]: import numpy as np
   
   In [2]: import pandas as pd
   
   In [3]: import re
   
   In [4]: df = pd.DataFrame({'col1':[1,2,3,4,5], 'col2':['admin', 'aa', 'bb', 'c_admin_d', 'ee_admin']})
   
   In [5]: df
   Out[5]: 
      col1       col2
   0     1      admin
   1     2         aa
   2     3         bb
   3     4  c_admin_d
   4     5   ee_admin
   
   In [6]: r = re.compile(r'.*(admin).*')
   
   In [7]: regmatch = np.vectorize(lambda x: bool(r.match(x)))
   
   In [8]: regmatch(df.col2.values)
   Out[8]: array([ True, False, False,  True,  True])
   
   In [9]: %timeit -n 100000 regmatch(df.col2.values)
   43.4 µs ± 362 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
   

Since you have changed your question to check any cell, and also concern about time efficiency:

# if you want to check all columns no mater what `dtypes` they are
dfs = df.astype(str, copy=True, errors='raise')
regmatch(dfs.values) # This will return a 2-d array of booleans
regmatch(dfs.values).any() # For existence.

You can still use df.applymap method, but again, it will be slower.

dfs = df.astype(str, copy=True, errors='raise')
r = re.compile(r'.*(admin).*')
dfs.applymap(lambda x: bool(r.match(x))) # This will return a dataframe of booleans.
dfs.applymap(lambda x: bool(r.match(x))).any().any() # For existence.