How to iterating string in df using python

Question

I have data

                     date                    id           request 
0     2016-06-17 09:25:05  [email protected]  GET HTTP/1.1   
1     2016-06-17 09:25:07  [email protected]     POST HTTP/1.1   
2     2016-06-17 09:25:47  [email protected]  CONNECT HTTP/1.1   
3     2016-06-17 09:25:47  [email protected]     POST HTTP/1.1   
4     2016-06-17 09:25:49  [email protected]  CONNECT HTTP/1.1

I need to iterate string and 'GET' not in df['request'] I want to delete string from df.

Desire output

               date                    id           request 
0     2016-06-17 09:25:05  [email protected]  GET HTTP/1.1

I try df = df['GET' in df.request] but it returns

KeyError: False

jezrael · Accepted Answer · 2016-06-23 12:15:40Z

2

You need boolean indexing with mask created by str.contains:

print (df.request.str.contains('GET'))
0  2016-06-17     True
1  2016-06-17    False
2  2016-06-17    False
3  2016-06-17    False
4  2016-06-17    False

print (df[df.request.str.contains('GET')])
                  date                    id       request
0 2016-06-17  09:25:05  [email protected]  GET HTTP/1.1

EDIT by comment:

For comparing column size use [], because size is function:

df_upd = df_upd[df_upd['size'].astype(int) > 3000]

edited Jun 23, 2016 at 12:15

answered Jun 23, 2016 at 12:03

jezrael

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6 Comments

NineWasps Over a year ago

Can you say, why df_upd = df_upd[int(df_upd.size) > 3000] return keyerror? I also need to compare size(I have this column), and if it less that 3000, delete that string

jezrael Over a year ago

You need astype for casting to int - df_upd = df_upd[df_upd.size.astype(int) > 3000]

NineWasps Over a year ago

it also return KeyError: True

jezrael Over a year ago

Hmmmm, there is another problem - size is function. So you need [] - df_upd = df_upd[df_upd['size'].astype(int) > 3000]

jezrael Over a year ago

btw, I have similar issues very often, because I like column names with same names as function like count, size...

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