I think you are looking for in:

if 'goat' in 'goat cheese':
    print('beeeeeeh!')

row['name'] is a dictionary with the value being the string you are searching. To search in a vectorized way, you don't need to use iterrows():

df['name'].str.contains('abc') will return a boolean index because it uses the pandas Series string contains method.

At the row level (if this is what you want) just use in:

for idx, row in df.iterrows():
      if 'abc' in row['name']:

Use numpy.where for a optimal vectorized solution, we don't need custom apply methods for such trivial actions:

df['col2'] = np.where(df['col1'].str.contains('A1'), 'Some Category', 'Another Category')
           # np.where(<condition>, <value if true>, <value if false>)
                    col1              col2
0       A1 zwd fill text     Some Category
1  B2 rest uninteresting  Another Category
2    A1 more random text     Some Category

Or purely pandas using Series.map:

df['col2'] = df['col1'].str.contains('A1').map({True: 'Some Category', 
                                                False: 'Another Category'})

                    col1              col2
0       A1 zwd fill text     Some Category
1  B2 rest uninteresting  Another Category
2    A1 more random text     Some Category

Timings:

# create test dataframe of 900k rows
df = pd.DataFrame({"col1": ["A1 zwd fill text", "B2 rest uninteresting", "A1 more random text"]})
dfbig = pd.concat([df]*300000, ignore_index=True)

Solution 1: np.where:

%%timeit
np.where(dfbig['col1'].str.contains('A1'), 'Some Category', 'Another Category')

855 ms ± 3.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Solution 2: Series.map:

%%timeit
dfbig['col1'].str.contains('A1').map({True: 'Some Category', 
                                                False: 'Another Category'})

920 ms ± 15.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Solution 3: apply:

%%timeit
dfbig.apply(lambda x: "Some Category" if "A1" in x.col1 else "Another Category", axis=1)

28.5 s ± 446 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Conclusion: numpy is ~135 x faster than apply

The error message states that you're trying to use .contains() on a string which is not a string function.

Try:

for i in range(0,len(final3)):
    default = final3["DefaultValue"].iloc[i]
    if not "|" in default:
        if default in final3["CodedData"].iloc[i]:
            final3["Condition"].iloc[i] = "False"
        else:
            final3["Condition"].iloc[i] = "True"

you can use python find() method

>>> s.apply(lambda x : any((i for i in x if i.find('you') >= 0)))
0    False
1     True
2    False
dtype: bool

I guess s.str.contains('you') is not working because elements of your series is not strings, but lists. But you can also do something like this:

>>> s.apply(lambda x: any(pd.Series(x).str.contains('you')))
0    False
1     True
2    False

Short answer: change data.columns=[headerName] into data.columns=headerName

Explanation: when you set data.columns=[headerName], the columns are MultiIndex object. Therefore, your log_df['Product'] is a DataFrame and for DataFrame, there is no str attribute.

When you set data.columns=headerName, your log_df['Product'] is a single column and you can use str attribute.

For any reason, if you need to keep your data as MultiIndex object, there is another solution: first convert your log_df['Product'] into Series. After that, str attribute is available.

products = pd.Series(df.Product.values.flatten())
include_clique = products[products.str.contains("Product A")]

However, I guess the first solution is what you're looking for

In [23]: df1
Out[23]: 
  Name    income
0    a        $1
1    b    $2,000
2    c   $10,000
3    d  $140,000

In [24]: cols_to_change = ['income']

In [25]: for col in cols_to_change:
    ...:     df1[col] = df1[col].str.replace('[$,]', '')
    ...: 

In [26]: df1
Out[26]: 
  Name  income
0    a       1
1    b    2000
2    c   10000
3    d  140000