I have a dataframe as follows:
Name Rating
0 ABC Good
1 XYZ Good #
2 GEH Good
3 ABH *
4 FEW Normal
Here I want to replace in the Rating element if it contain # it should replace by Can be improve , if it contain * then Very Poor. I have tried with following but it replace whole string. But I want to replace only the special char if it present.But it solves for another case if only special char is present.
import pandas as pd
df = pd.DataFrame() # Load with data
df['Rating'] = df['Rating'].str.replace('.*#+.*', 'Can be improve')
is returning
Name Rating
0 ABC Good
1 XYZ Can be improve
2 GEH Good
3 ABH Very Poor
4 FEW Normal
Can anybody help me out with this?
import pandas as pd
df = pd.DataFrame({"Rating": ["Good", "Good #", "*"]})
df["Rating"] = df["Rating"].str.replace("#", "Can be improve")
df["Rating"] = df["Rating"].str.replace("*", "Very Poor")
print(df)
Output:
0 Good
1 Good Can be improve
2 Very Poor
You replace the whole string because .* matches any character zero or more times.
If your special values are always at the end of the string you might use:
.str.replace(r'#$', "Can be improve")
.str.replace(r'\*$', "Very Poor")
Related
I would like to keep in the column only the two first words of a cell in a dataframe.
For instance:
df = pd.DataFrame(["I'm learning Python", "I don't have money"])
I would like that the results in the column have the following output:
"I'm learning" ; "I don't"
After that, if possible I would like to add '*' between each word. So would be like:
"*I'm* *learning*" ; "*I* *don't*"
Thanks for all the help!
You can use a regex with str.replace:
df[0].str.replace(r'(\S+)\s(\S+).*', r'*\1* *\2*', regex=True)
output:
0 *I'm* *learning*
1 *I* *don't*
Name: 0, dtype: object
As a new column:
df['new'] = df[0].str.replace(r'(\S+)\s(\S+).*', r'*\1* *\2*', regex=True)
output:
0 new
0 I'm learning Python *I'm* *learning*
1 I don't have money *I* *don't*
I'm trying to change the Strings "SLL" under the competitions column to "League" but when i tried this:
messi_dataset.replace("SLL", "League",regex = True)
It only changed the first "SLL" to "League" but then other strings that were "SLL" became "UCL. I have no idea why. I also tried changing regex = True to inlace = True but no luck.
https://drive.google.com/file/d/1ldq6o70j-FsjX832GbYq24jzeR0IwlEs/view?usp=sharing
https://drive.google.com/file/d/1OeCSutkfdHdroCmTEG9KqnYypso3bwDm/view?usp=sharing
Suppose you have a dataframe as below:
import pandas as pd
import re
df = pd.DataFrame({'Competitions': ['SLL', 'sll','apple', 'banana', 'aabbSLL', 'ccddSLL']})
# write a regex pattern that replaces 'SLL'
# I assumed case-irrelevant
regex_pat = re.compile(r'SLL', flags=re.IGNORECASE)
df['Competitions'].str.replace(regex_pat, 'league', regex=True)
# Input DataFrame
Competitions
0 SLL
1 sll
2 apple
3 banana
4 aabbSLL
5 ccddSLL
Output:
0 league
1 league
2 apple
3 banana
4 aabbleague
5 ccddleague
Name: Competitions, dtype: object
Hope it clarifies.
base on this Answer test this code:
messi_dataset['competitions'] = messi_dataset['competitions'].replace("SLL", "League")
also, there are many different ways to do this like this one that I test:
messi_dataset.replace({'competitions': 'SLL'}, "League")
for those cases that 'SLL' is a part of another word:
messi_dataset.replace({'competitions': 'SLL'}, "League", regex=True)
I have entire table as string like below:
a= "id;date;type;status;description\r\n1;20-Jan-2019;cat1;active;customer is under\xe9e observation\r\n2;18-Feb-2019;cat2;active;customer is genuine\r\n"
inside string we do have some ascii code like \xe9e so we have to convert the string to non-ascii
My expected output is to convert above string to a dataframe
as below:
id date type status description
1 20-Jan-2019 cat1 active customer is under observation
2 18-Feb-2019 cat2 active customer is genuine
My code :
b = a.splitlines()
c = pd.DataFrame([sub.split(";") for sub in b])
I am getting the following output. but I need the fist row as my header and also convert the ascii to utf-8 text.
0 1 2 3 4 5 6
0 id date type status description None None
1 1 20-Jan-2019 cat1 active customer is underée observation None None
2 2 18-Feb-2019 cat2 active customer is genuine None None
Also, please not here it is creating extra columns with value None. Which should not be the case
Here is a bit of a hacky answer, but given that your question isn't really clear, this should hopefully be sufficient.
import pandas as pd
import numpy as np
import re
a="id;date;type;status;description\r\n1;20-Jan-2019;cat1;active;customer is under\xe9e observation\r\n2;18-Feb-2019;cat2;active;customer is genuine\r\n"
b=re.split('; |\r|\n',a) #split at the delimiters.
del b[-1] #also delete the last index, which we dont need
b[1:]=[re.sub(r'\xe9e', '', b[i]) for i in range(1,len(b))] #get rid of that \xe9e issue
df=pd.DataFrame([b[i].split(';') for i in range(1,len(b))]) #make the dataframe
##list comprehension allows to generalize this if you add to string##
df.columns=b[0].split(';') #split the title words for column names
df['id']=[i for i in range(1,len(b))]
df
This output is presumably what you meant by a dataframe:
Looking for some help.
I have a pandas dataframe column and I want to extract the prefix where such prefix exists in a separate list.
pr_list = ['1 FO-','2 IA-']
Column in df is like
PartNumber
ABC
DEF
1 FO-BLABLA
2 IA-EXAMPLE
What I am looking for is to extract the prefix where present, put in a new column and leave the rest of the string in the original column.
PartNumber Prefix
ABC
DEF
BLABLA 1 FO-
EXAMPLE 2 IA-
Have tried some things like str.startswith but a bit of a python novice and wasn't able to get it to work.
much appreciated
EDIT
Both solutions below work on the test data, however I am getting an error
error: nothing to repeat at position 16
Which suggests something askew in my dataset. Not sure what position 16 refers to but looking at both the prefix list and PartNumber column in position 16 nothing seems out of the ordinary?
EDIT 2
I have traced it to have an * in the pr_list seems to be throwing it. is * some reserved character? is there a way to break it out so it is read as text?
You can try:
df['Prefix']=df.PartNumber.str.extract(r'({})'.format('|'.join(pr_list))).fillna('')
df.PartNumber=df.PartNumber.str.replace('|'.join(pr_list),'')
print(df)
PartNumber Prefix
0 ABC
1 DEF
2 BLABLA 1 FO-
3 EXAMPLE 2 IA-
Maybe it's not what you are looking for, but may it help.
import pandas as pd
pr_list = ['1 FO-','2 IA-']
df = pd.DataFrame({'PartNumber':['ABC','DEF','1 FO-BLABLA','2 IA-EXAMPLE']})
extr = '|'.join(x for x in pr_list)
df['Prefix'] = df['PartNumber'].str.extract('('+ extr + ')', expand=False).fillna('')
df['PartNumber'] = df['PartNumber'].str.replace('|'.join(pr_list),'')
df
I have the following data frame (consisting of both negative and positive numbers):
df.head()
Out[39]:
Prices
0 -445.0
1 -2058.0
2 -954.0
3 -520.0
4 -730.0
I am trying to change the 'Prices' column to display as currency when I export it to an Excel spreadsheet. The following command I use works well:
df['Prices'] = df['Prices'].map("${:,.0f}".format)
df.head()
Out[42]:
Prices
0 $-445
1 $-2,058
2 $-954
3 $-520
4 $-730
Now my question here is what would I do if I wanted the output to have the negative signs BEFORE the dollar sign. In the output above, the dollar signs are before the negative signs. I am looking for something like this:
-$445
-$2,058
-$954
-$520
-$730
Please note there are also positive numbers as well.
You can use np.where and test whether the values are negative and if so prepend a negative sign in front of the dollar and cast the series to a string using astype:
In [153]:
df['Prices'] = np.where( df['Prices'] < 0, '-$' + df['Prices'].astype(str).str[1:], '$' + df['Prices'].astype(str))
df['Prices']
Out[153]:
0 -$445.0
1 -$2058.0
2 -$954.0
3 -$520.0
4 -$730.0
Name: Prices, dtype: object
You can use the locale module and the _override_localeconv dict. It's not well documented, but it's a trick I found in another answer that has helped me before.
import pandas as pd
import locale
locale.setlocale( locale.LC_ALL, 'English_United States.1252')
# Made an assumption with that locale. Adjust as appropriate.
locale._override_localeconv = {'n_sign_posn':1}
# Load dataframe into df
df['Prices'] = df['Prices'].map(locale.currency)
This creates a dataframe that looks like this:
Prices
0 -$445.00
1 -$2058.00
2 -$954.00
3 -$520.00
4 -$730.00