Replace 'space' in certain value of csv columns-CodePudding

I've got the next csv file:

Summary,Issue key,Issue id,Issue Type,Status,Project key,Attachment,Attachment.1,Attachment.2,Attachment.3,Attachment.4,Attachment.5

Find issue,IS-11,576,Task,Solved,One-1,10/28/21 11:49;Olga_Sokolova;SALUPRJBKK-1663_2021-10-28 14-38-01-372.mp4;file://SALUPRJBKK/SALUPRJBKK-1663/SALUPRJBKK-1663_2021-10-28 14-38-01-372.mp4

I need to choose all the attachments values and replace the "space" in the filename to " ". The main problem is to skip the first 'space' after the date in the attachment value and also to get all the attachment value. I tried to use standart csv reader, pandas and etc. but I can only get name of the column

import pandas as pd

data = pd.read_csv("SALUPRJBKK_new_10.csv")
for i in data:
    if "Attachment" in i:
        print(i)

CodePudding user response：

Select 'Attachment' columns using filter and replace all whitespaces by ' ' then update your dataframe in place:

df.update(df.filter(like='Attachment').replace(' ', ' ', regex=True))

My advise if you need to escape HTML entities is to use quote from urllib module:

from urllib.parse import quote

df.update(df.filter(like='Attachment').fillna('').applymap(quote))

CodePudding user response：

consider specifying the separotor:

data = pd.read_csv("SALUPRJBKK_new_10.csv", sep=",")

Note: there is a mix of seperator in your example "," or ";"

and if you want to acces only to "Attachment" do:

data["Attachment"]

CodePudding user response：

I think do you need a url encode.

Try with this:

import urllib.parse

query = 'Hellóóó W r l d @ Pyt@@ h on.mp4'
newUrl = urllib.parse.quote(query)
print(newUrl)

And now the encode result:

Hellóóó W r l d @ Pyt@@ h on.mp4

The text or url is encoded and change all the special characters in string.