How do I remove the rest of a string after a character in a dataframe column?

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I have a dataframe that contains userdata. There is a column that includes filenames that users have accessed. The filenames look like this:

blah-blah-blah/dss_outline.pdf  
doot-doot/helper_doc.pdf
blah-blah-blah/help_file.pdf

My goal is to chop off everything after and including the / so that I can just look at the top-level programs people are examining (which the numerous different files are organized under).

So, I'm having two challenges:

1 - How do I 'grab' everything up to the '/'? I've been looking at regex, but I'm having a hard time writing the correct expression.

2 - How do I replace all of the filenames with the concatenated filename? I found that I could use df['Filename'] = df['Filename'].str.split('/')[0] to grab the proper portion, but it won't apply across the series object. That's the logic of what I want to do, but I can't figure out how to do it.

Thanks

You have lot of solutions handy:

1) Just with split() method:
>>> df
                             col1
0  blah-blah-blah/dss_outline.pdf
1        doot-doot/helper_doc.pdf
2    blah-blah-blah/help_file.pdf


>>> df['col1'].str.split('/', 1).str[0].str.strip()
0    blah-blah-blah
1         doot-doot
2    blah-blah-blah

Name: col1, dtype: object

2) You can use apply() + split()
>>> df['col1'].apply(lambda s: s.split('/')[0])
0    blah-blah-blah
1         doot-doot
2    blah-blah-blah
Name: col1, dtype: object
3) You can use rsplit() + str[0] to strip off the desired:
>>> df['col1'].str.rsplit('/').str[0]
0    blah-blah-blah
1         doot-doot
2    blah-blah-blah
Name: col1, dtype: object
4) You can use pandas native regex With extract():
>>> df['col1'] = df['col1'].str.extract('([^/]+)')
>>> df
             col1
0  blah-blah-blah
1       doot-doot
2  blah-blah-blah

OR
# df.col1.str.extract('([^/]+)')

Pandas delete parts of string after specified character inside a , You can reformat the values by passing a reformatting function into the apply method as follows: from StringIO import StringIO import pandas as pd data = """ obs  When you wish to remove the character by using its code. This can help you in removing case sensitive character. Just use the char (code) in place of remove_char. To know the code of the character uses the function shown below.

You may use \/.*$ to match the part you don't need and remove it: DEMO This matches a forward slash and any following character till the end of the string (be careful to use a multiline flag if your engine needs it!).

OR you may use ^[^/]+ to match the part you want and extract it: DEMO This matches any consecutive characters except / from the beginning of a string (again, multiline needed!).

How do I remove the rest of a string after a character in a dataframe , You have lot of solutions handy: 1) Just with split() method: >>> df col1 0 blah- blah-blah/dss_outline.pdf 1 doot-doot/helper_doc.pdf 2  Remove Text after a specific character using FIND&Select command You can also use the Find and Replace command to remove text after a specified character, just refer to the following steps: 1# Click “ HOME “->” Find&Select ”->” Replace… ”, then the window of the Find and Replace will appear.

Use series.apply():

>>> import pandas
>>> data = {'filename': ["blah-blah-blah/dss_outline.pdf", "doot-doot/helper_doc.pdf", "blah-blah-blah/help_file.pdf"]}
>>> df = pandas.DataFrame(data=data)
>>> df
                         filename
0  blah-blah-blah/dss_outline.pdf
1        doot-doot/helper_doc.pdf
2    blah-blah-blah/help_file.pdf
>>> def get_top_level_from(string):
...     return string.split('/')[0]
... 
>>> series = df["filename"]
>>> series
0    blah-blah-blah/dss_outline.pdf
1          doot-doot/helper_doc.pdf
2      blah-blah-blah/help_file.pdf
Name: filename, dtype: object
>>> series.apply(get_top_level_from)
0    blah-blah-blah
1         doot-doot
2    blah-blah-blah
Name: filename, dtype: object

Code:

def get_top_level_from(string):
    return string.split('/')[0]

results = df["filename"].apply(get_top_level_from)

Removing characters before, after, and in the middle of strings, Removing characters before, after, and in the middle of strings and pandas, you will need to remove characters from your strings *a lot*. Below is the solution in one line of code, after that, I'll break it down step by step. I am able to remove the values for one specific column, however the split() function does not apply to the whole dataframe. f = lambda x: x["Australia"].split(" ")[0] df = df.apply(f, axis=1) Anyone an idea how I could remove the information after a space occures for each value in the dataframe?

Use df.replace

df.replace('\/.*$','',regex=True)


              col
0  blah-blah-blah
1       doot-doot
2  blah-blah-blah

8 ways to apply LEFT, RIGHT, MID in Pandas, At times, you may need to extract specific characters within a string. DataFrame (Data, columns= ['Identifier']) Left = df['Identifier'].str[:5] print (Left) You may also face situations where you'd like to get all the characters after a symbol (the  RIGHT + LEN. Using REPLACE. Combine MID and LEN. Text to Column. UDF. 1. Combine RIGHT and LEN to Remove the First Character from the Value. Using a combination of RIGHT and LEN is the most suitable way to remove the first character from a cell or fr.

Split a column in Pandas dataframe and get part of it, str accessor, it does fast vectorized string operations for Series and Dataframes and returns a string object. Pandas str accessor has number of useful methods and  Since our example vector is shorter than 1000000L, the whole rest of the vector after position 7 is printed. Example 3: Replace Substring with substr() & substring() Another popular usage of the substr and substring R functions is the replacement of certain characters in a string. This is again something we can do with both functions.

Python, Python break statement · Python Continue Statement · Python pass Statement · Looping If no uppercase characters exist, it returns the original string. DataFrame(data) # converting and overwriting values in column df["Name"]= As shown in the output image, the comparison is true after removing the left side spaces  String split the column of dataframe in pandas python: String split can be achieved in two steps (i) Convert the dataframe column to list and split the list (ii) Convert the splitted list into dataframe. Step 1: Convert the dataframe column to list and split the list: df1.State.str.split().tolist()

pandas.Series.str.replace, String can be a character sequence or regular expression. replstr or callable. Replacement string or a callable. The callable is passed the regex match object and  Since in our example the ‘DataFrame Column’ is the Price column (which contains the strings values), you’ll then need to add the following syntax: df['Price'] = df['Price'].astype(int) So this is the complete Python code that you may apply to convert the strings into integers in the pandas DataFrame:

Comments
  • "but it won't apply across the series object" meaning? can you please explain any further. :)
  • df['your_column_name'] = df['your_column_name'].map(lambda a: os.path.basename(a)) Depending on your folder structure, as suggested below you could use str.split() as well
  • This is excellent, thank you for sharing the numerous options here! I'm still getting up to speed on pandas/regex/lambda expressions so it's nice to see them all represented here.
  • @RNGeezy, thnx for liking and accepting this as an answer, happy learning :-)
  • How do I apply that across the column in the dataframe? So, say I used re.sub() to replace the unwanted bits with an empty string, how do I apply that row by row?
  • You will have to iterate over the rows and apply it one by one, i guess.