import pandas as pd

# Create Dataframe having 10 rows and 2 columns 'code' and 'URL'
df = pd.DataFrame({'code': [1,1,2,2,3,4,1,2,2,5],
                   'URL': ['www.abc.de','https://www.abc.fr/-de','www.abc.fr','www.abc.fr','www.abc.co.uk','www.abc.es','www.abc.de','www.abc.fr','www.abc.fr','www.abc.it']})

# Create new dataframe by filtering out all rows where the column 'code' is equal to 1
new_df = df[df['code'] == 1]

# Below is how the new dataframe looks like
print(new_df)

                      URL  code
0              www.abc.de     1
1  https://www.abc.fr/-de     1
6              www.abc.de     1

Below are the dtypes for reference
print(new_df.dtypes)
URL     object
code     int64
dtype: object

# Now I am trying to exclude all those rows where the 'URL' column does not have .de as the pattern. This should retain only the 2nd row in new_df from above output
new_df = new_df[~ new_df['URL'].str.contains(r".de", case = True)]

# Below is how the output looks like
print(new_df)
Empty DataFrame
Columns: [URL, code]
Index: []

new_df <- new_df[grep(".de",new_df$URL, fixed = TRUE, invert = TRUE), ]

# Desired output for new_df
                   URL  code
https://www.abc.fr/-de     1