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

This could be because your dataframe has multiple columns with the same name. I can recreate this error by doing the following:

import pandas as pd

df = pd.DataFrame([['0','1','2'],['2','4','5']],columns = ['a','b','b'])
for c in df.columns:
    print(df[c].str.replace("1",""))

The problem is that once you get to the repeated column name (in my example, when c == 'b'), then df[c] is actually a dataframe with 2 cols and .str is not available.

So if this is your issue, find the columns with the same name and give them unique names.

Also, as mentioned by @JonClements it isn't necessary to loop over the columns at all, you can just do df = df.replace('.*0.*', '', regex=True)

The error means exactly what it says:

AttributeError: 'str' object has no attribute 'DataFrame' 
      ^           ^                                ^
the kind of error |                                |
       the thing you tried to use      what was missing from it

The line it's complaining about:

df = pd.DataFrame(date, columns = ['Date'])
     ^      ^
     |   the attribute the error said was missing
the thing the error said was a string

has been working no problem until I added a few lines of code above

Evidently, somewhere in the "few lines of code above", you caused pd to be a string. And sure enough, when we look at those few lines of code, we find:

pd = PDays[j]
^       ^
|    the string that you're making it into
the thing that you're making a string

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

The function pd.read_csv() is already a DataFrame and thus that kind of object does not support calling .to_dataframe().

You can check the type of your variable ds using print(type(ds)), you will see that it is a pandas DataFrame type.

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

First problem shoud be duplicated columns names, so after select colB get not Series, but DataFrame:

df = pd.DataFrame([['Example: s', 'as', 2], ['dd', 'aaa', 3]], columns=['colB','colB','colC'])
print (df)
         colB colB  colC
0  Example: s   as     2
1          dd  aaa     3

print (df['colB'])
         colB colB
0  Example: s   as
1          dd  aaa

#print (df['colB'].str.contains('Example:'))
#>AttributeError: 'DataFrame' object has no attribute 'str'

Solution should be join columns together:

print (df['colB'].apply(' '.join, axis=1))
0    Example: s as
1           dd aaa

df['colB'] = df.pop('colB').apply(' '.join, axis=1)
df = df[~df['colB'].str.contains('Example:')] 
print (df)
   colC    colB
1     3  dd aaa

Second problem should be hidden MultiIndex:

df = pd.DataFrame([['Example: s', 'as', 2], ['dd', 'aaa', 3]], columns=['colA','colB','colC'])
df.columns = pd.MultiIndex.from_arrays([df.columns])
print (df)
         colA colB colC
0  Example: s   as    2
1          dd  aaa    3

print (df['colB'])
  colB
0   as
1  aaa

#print (df['colB'].str.contains('Example:'))
#>AttributeError: 'DataFrame' object has no attribute 'str'

Solution is reassign first level:

df.columns = df.columns.get_level_values(0)
df = df[~df['colB'].str.contains('Example:')] 
print (df)
         colA colB  colC
0  Example: s   as     2
1          dd  aaa     3

And third should be MultiIndex:

df = pd.DataFrame([['Example: s', 'as', 2], ['dd', 'aaa', 3]], columns=['colA','colB','colC'])
df.columns = pd.MultiIndex.from_product([df.columns, ['a']])
print (df)
         colA colB colC
            a    a    a
0  Example: s   as    2
1          dd  aaa    3

print (df['colB'])
     a
0   as
1  aaa

print (df.columns)
MultiIndex(levels=[['colA', 'colB', 'colC'], ['a']],
           codes=[[0, 1, 2], [0, 0, 0]])

#print (df['colB'].str.contains('Example:'))
#>AttributeError: 'DataFrame' object has no attribute 'str'

Solution is select MultiIndex by tuple:

df1 = df[~df[('colB', 'a')].str.contains('Example:')] 
print (df1)
         colA colB colC
            a    a    a
0  Example: s   as    2
1          dd  aaa    3

Or reassign back:

df.columns = df.columns.get_level_values(0)
df2 = df[~df['colB'].str.contains('Example:')] 
print (df2)
         colA colB  colC
0  Example: s   as     2
1          dd  aaa     3

Or remove second level:

df.columns = df.columns.droplevel(1)
df2 = df[~df['colB'].str.contains('Example:')] 
print (df2)
         colA colB  colC
0  Example: s   as     2
1          dd  aaa     3

You need to use apply:

DataFrame[['ColumnA','ColumnB']].apply(lambda s: s.str.split('\:').str[-1].str.strip())

Or a list comprehension, which can be significantly faster as chaining str methods can be slow:

[[s.split('\:')[-1].strip() for s in df[col]] for col in ['ColumnA','ColumnB']]