Develop Reference

How do I print out the Phone number from a csv with padded 0 using pandas?

So I have a CSV file with the following content:
Person,Phone
One,08001111111
Two,08002222222
Three,08003333333
When I used the following code:
import pandas as pd
df = pd.read_csv('test_stuff.csv')
print(df)
It prints out:
Person Phone
0 One 8001111111
1 Two 8002222222
2 Three 8003333333
It removed the starting 0 from the Phone column. I then tried to add the phone as string in the csv file, like so:
Person,Phone
One,'08001111111'
Two,'08002222222'
Three,'08003333333'
However, the result is now this:
Person Phone
0 One '08001111111'
1 Two '08002222222'
2 Three '08003333333'
What can I do to resolve this? I am hoping for a result like this:
Person Phone
0 One 08001111111
1 Two 08002222222
2 Three 08003333333
Thanks in advance.

Don't try to add the zeros back, just don't delete them in the first place by telling pandas.read_csv that you column is a string:
pd.read_csv('test_stuff.csv', dtype={'Phone': 'str'})
output:
Person Phone
0 One 08001111111
1 Two 08002222222
2 Three 08003333333

strings to column using python

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:

Extract prefix from string in dataframe column where exists in a list

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

Parsing a JSON string enclosed with quotation marks from a CSV using Pandas

Similar to this question, but my CSV has a slightly different format. Here is an example:
id,employee,details,createdAt
1,John,"{"Country":"USA","Salary":5000,"Review":null}","2018-09-01"
2,Sarah,"{"Country":"Australia", "Salary":6000,"Review":"Hardworking"}","2018-09-05"
I think the double quotation mark in the beginning of the JSON column might have caused some errors. Using df = pandas.read_csv('file.csv'), this is the dataframe that I got:
id employee details createdAt Unnamed: 1 Unnamed: 2
1 John {Country":"USA" Salary:5000 Review:null}" 2018-09-01
2 Sarah {Country":"Australia" Salary:6000 Review:"Hardworking"}" 2018-09-05
My desired output:
id employee details createdAt
1 John {"Country":"USA","Salary":5000,"Review":null} 2018-09-01
2 Sarah {"Country":"Australia","Salary":6000,"Review":"Hardworking"} 2018-09-05
I've tried adding quotechar='"' as the parameter and it still doesn't give me the result that I want. Is there a way to tell pandas to ignore the first and the last quotation mark surrounding the json value?

As an alternative approach you could read the file in manually, parse each row correctly and use the resulting data to contruct the dataframe. This works by splitting the row both forward and backwards to get the non-problematic columns and then taking the remaining part:
import pandas as pd
data = []
with open("e1.csv") as f_input:
for row in f_input:
row = row.strip()
split = row.split(',', 2)
rsplit = [cell.strip('"') for cell in split[-1].rsplit(',', 1)]
data.append(split[0:2] + rsplit)
df = pd.DataFrame(data[1:], columns=data[0])
print(df)
This would display your data as:
id employee details createdAt
0 1 John {"Country":"USA","Salary":5000,"Review":null} 2018-09-01
1 2 Sarah {"Country":"Australia", "Salary":6000,"Review"... 2018-09-05

I have reproduced your file
With
df = pd.read_csv('e1.csv', index_col=None )
print (df)
Output
id emp details createdat
0 1 john "{"Country":"USA","Salary":5000,"Review":null}" "2018-09-01"
1 2 sarah "{"Country":"Australia", "Salary":6000,"Review... "2018-09-05"

I think there's a better way by passing a regex to sep=r',"|",|(?<=\d),' and possibly some other combination of parameters. I haven't figured it out totally.
Here is a less than optimal option:
df = pd.read_csv('s083838383.csv', sep='##$%^', engine='python')
header = df.columns[0]
print(df)
Why sep='##$%^' ? This is just garbage that allows you to read the file with no sep character. It could be any random character and is just used as a means to import the data into a df object to work with.
df looks like this:
id,employee,details,createdAt
0 1,John,"{"Country":"USA","Salary":5000,"Review...
1 2,Sarah,"{"Country":"Australia", "Salary":6000...
Then you could use str.extract to apply regex and expand the columns:
result = df[header].str.extract(r'(.+),(.+),("\{.+\}"),(.+)',
expand=True).applymap(str.strip)
result.columns = header.strip().split(',')
print(result)
result is:
id employee details createdAt
0 1 John "{"Country":"USA","Salary":5000,"Review":null}" "2018-09-01"
1 2 Sarah "{"Country":"Australia", "Salary":6000,"Review... "2018-09-05"
If you need the starting and ending quotes stripped off of the details string values, you could do:
result['details'] = result['details'].str.strip('"')
If the details object items needs to be a dicts instead of strings, you could do:
from json import loads
result['details'] = result['details'].apply(loads)

Extracting many URLs in a python dataframe

I have a dataframe which contains text including one or more URL(s) :
user_id text
1 blabla... http://amazon.com ...blabla
1 blabla... http://nasa.com ...blabla
2 blabla... https://google.com ...blabla ...https://yahoo.com ...blabla
2 blabla... https://fnac.com ...blabla ...
3 blabla....
I want to transform this dataframe with the count of URL(s) per user-id :
user_id count_URL
1 2
2 3
3 0
Is there a simple way to perform this task in Python ?
My code start :
URL = pd.DataFrame(columns=['A','B','C','D','E','F','G'])
for i in range(data.shape[0]) :
for j in range(0,8):
URL.iloc[i,j] = re.findall("(?P<url>https?://[^\s]+)", str(data.iloc[i]))
Thanks you
Lionel

In general, the definition of a URL is much more complex than what you have in your example. Unless you are sure you have very simple URLs, you should look up a good pattern.
import re
URLPATTERN = r'(https?://\S+)' # Lousy, but...
First, extract the URLs from each string and count them:
df['urlcount'] = df.text.apply(lambda x: re.findall(URLPATTERN, x)).str.len()
Next, group the counts by user id:
df.groupby('user_id').sum()['urlcount']
#user_id
#1 2
#2 3
#3 0

Below there is another way to do that:
#read data
import pandas as pd
data = pd.read_csv("data.csv")
#Divide data into URL and user_id and cast it to pandas DataFrame
URL = pd.DataFrame(data.loc[:,"text"].values)
user_id = pd.DataFrame(data.loc[:,"user_id"].values)
#count the number of appearance of the "http" in each row of data
sub = "http"
count_URL = []
for val in URL.iterrows():
counter = val[1][0].count(sub)
count_URL.append(counter)
#list to DataFrame
count_URL = pd.DataFrame(count_URL)
#Concatenate the two data frames and apply the code of #DyZ to group by and count the number of url
finalDF = pd.concat([user_id,count_URL],axis=1)
finalDF.columns=["user_id","urlcount"]
data = finalDF.groupby('user_id').sum()['urlcount']
print(data.head())