get non numerical rows in a column pandas python

Use boolean indexing with mask created by to_numeric + isnull
Note: This solution does not find or filter numbers saved as strings: like '1' or '22'

print (pd.to_numeric(df['num'], errors='coerce'))
0   -1.48
1    1.70
2   -6.18
3    0.25
4     NaN
5    0.25
Name: num, dtype: float64

print (pd.to_numeric(df['num'], errors='coerce').isnull())
0    False
1    False
2    False
3    False
4     True
5    False
Name: num, dtype: bool

print (df[pd.to_numeric(df['num'], errors='coerce').isnull()])
  N-D     num unit
4  Q5  sum(d)   UD

Another solution with isinstance and apply:

print (df[df['num'].apply(lambda x: isinstance(x, str))])
  N-D     num unit
4  Q5  sum(d)   UD

Old topic, but if the numbers have been converted to 'str', type(x) == str is not working.

Instead, it's better to use isnumeric() or isdigit().

df = df[df['num'].apply(lambda x: not x.isnumeric())]

I tested all three approaches on my own dataframe with 200k+ rows, assuming numbers have been converted to 'str' by pd.read_csv().

def f1():
    df[pd.to_numeric(df['num'], errors='coerce').isnull()]

def f2():
    df[~df.num.str.match('^\-?(\d*\.?\d+|\d+\.?\d*)$')]

def f3():
    df[df['num'].apply(lambda x: not x.isnumeric())]

I got following execution times by running each function 10 times.

timeit.timeit(f1, number=10)
1.04128568888882

timeit.timeit(f2, number=10)
1.959099448888992

timeit.timeit(f3, number=10)
0.48741375999998127

Conculsion: fastest method is isnumeric(), slowest is regular expression method.

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Edit: As @set92 commented, isnumeric() works for integer only. So the fastest applicable function is pd.to_numeric() to have a universal solutions works for any type of numerical values.

It is possible to define a isfloat() function in python; but it will be slower than internal functions, especially for big DataFrames.

tmp=['4.0','4','4.5','1','test']*200000
df=pd.DataFrame(data=tmp,columns=['num'])


def f1():
    df[pd.to_numeric(df['num'], errors='coerce').isnull()]

def f2():
    df[df['num'].apply(lambda x: not isfloat(x))] 

def f3():
    df[~df.num.str.match('^\-?(\d*\.?\d+|\d+\.?\d*)$')]


print('to_numeric:',timeit.timeit(f1, number=10))
print('isfloat:',timeit.timeit(f2, number=10))
print('regular exp:',timeit.timeit(f3, number=10))

Results:

to_numeric: 8.303612694763615
isfloat: 9.972200270603594
regular exp: 11.420604273894583