Develop Reference

querying a multiindex pandas dataframe with slices

Assuming I have the following multiindex DF
import pandas as pd
import numpy as np
import pandas as pd
input_id = np.array(['12345'])
docType = np.array(['pre','pub','app','dw'])
docId = np.array(['34455667'])
sec_type = np.array(['bib','abs','cl','de'])
sec_ids = np.array(['x-y','z-k'])
index = pd.MultiIndex.from_product([input_id,docType,docId,sec_type,sec_ids])
content= [str(randint(1,10))+ '##' + str(randint(1,10)) for i in range(len(index))]
df = pd.DataFrame(content, index=index, columns=['content'])
df.rename_axis(index=['input_id','docType','docId','secType','sec_ids'], inplace=True)
df
I know that I can query a multiindex DF as follows:
# querying a multiindex DF
idx = pd.IndexSlice
df.loc[idx[:,['pub','pre'],:,'de',:]]
basically with the help of pd.IndexSlice I can pass the values I want for every of the indexes. In the above case I want the resulting DF where the second index is 'pub' OR 'pre' and the 4th one is 'de'.
I am looking for the way to pass a range of values to the query. something like multiindex 3 beeing between 34567 and 45657. Assume those are integers.
pseudocode: df.loc[idx[:,['pub','pre'],XXXXX,'de',:]]
XXXX = ?
EDIT 1:
docId column index is of text type, probably its necessary to change it first to int

Turns out query is very powerful:
df.query('docType in ["pub","pre"] and ("34455667" <= docId <= "3445568") and (secType=="de")')
Output:
content
input_id docType docId secType sec_ids
12345 pre 34455667 de x-y 2##9
z-k 6##1
pub 34455667 de x-y 6##5
z-k 9##8

Pandas read_csv for a no quote file

I'm trying to read a file that doesn't have any quotes, which is causing inconsistent number of row lengths
Data looks as follows:
col_a, col_b
abc, inc., 5
xyz corb, 10
Since there are no quotes around "abc, inc.", this is causing the first row to get split into 3 values, but it should actually be just 2 values.
This column is not necessarily in the first position, and that there can be another bad column like this. The data has around 250 columns.
I'm reading this using pd.read_csv, how can this be resolved?
Thanks!

Its not a CSV but since there is only one column with the errant commas you can process with the csv module and fix the slice that holds too many column values. When a row has too many cells, assume they are the ones from the unescaped comma.
import pandas as pd
import csv
def split_badrows(fileobj, bad_col, total_cols):
"""Iterate rows, colapsing extra columns at bad_col"""
for row in csv.reader(fileobj):
row = [cell.strip() for cell in row]
extras = len(row) - total_cols
if extras > 0:
# colapse slice at troubled column into single value
extras += 1 # python slice doesn't include right endpoint
row[bad_col] = ", ".join(row[bad_col:bad_col+extras])
del row[bad_col+1:bad_col+extras]
yield row
def df_from_badtext(fileobj, bad_col):
"""Make pandas.DataFrame from badly formatted text"""
columns = [cell.strip() for cell in next(fileobj).split(",")]
total_cols = len(columns)
return pd.DataFrame(split_badrows(fileobj, bad_col, total_cols),
columns=columns)
# test
open("testme.txt", "w").write("""col_a, col_b
abc, inc., 5
xyz corb, 10""")
df = df_from_badtext(open("testme.txt"), bad_col=0)
print(df)

Data split to list then transform to dataframe.
csv = '''col_a, col_b
abc, inc., 5
xyz corb, 10'''+'\n'
import re
import pandas as pd
reArr = re.findall('(.*),([^,]+)\n',csv)
df=pd.DataFrame(reArr[1:],columns=reArr[0])
print(df)
col_a
col_b
0
abc, inc.
5
1
xyz corb
10

EDIT:
Thanks to tdelaney comment below:
see if this works
pd.read_csv('foo.csv',delimiter=",(?!( [\w\d]*).,)").dropna(axis=1)
OLD:
using delimiter as ",(?!.*,)" in read_csv seems to be solving this for me

EDIT (after updated question with an additional column):
Solution 1:
You can create a function with the bad column as a parameter and use split and concat to correct the dataframe depending on that bad column. Please note that the bad_col parameter in my function is the column number, where we start counting at 1, rather than 0 (e.g. 1, 2, 3, etc. instead of 0, 1, 2, etc.):
import pandas as pd
import numpy as np
from io import StringIO
data = StringIO('''
col, col_a, col_b
000, abc, inc., 5
111, xyz corb, 10
''')
df = pd.read_csv(data, sep="|")
def fix_csv(df, bad_col):
cols = df.columns.str.split(', ')[0]
x = len(cols) - bad_col
tmp = df.iloc[:,0].str.split(', ', expand=True, n=x)
df = pd.concat([tmp.iloc[:,0],
tmp.iloc[:,-1].str.rsplit(', ', expand=True, n=x)],
axis=1)
df.columns = cols
return df
fix_csv(df, bad_col=2)
Solution 2 (this is if you have issues in multiple columns and you need to use more brute force):
It sounds like there is a possibility that you there could be multiple columns affected from the comments as you mentioned only 1 "so far".
As such, this might be a little bit of a project to clean up the data. The following code can give you an idea how to do that. The bottom-line is that you can create two different dataframes: 1) The first dataframe has the minimum number of commas (i.e. they should be the rows without any issues). 2) The other dataframe will be the dataframe with all of the issues. I've shown how you can clean the data to get to the correct number of columns and then change the data back and concat the two dataframes.
import pandas as pd
import numpy as np
from io import StringIO
data = StringIO('''
col, col_a, col_b
000, abc, inc., 5
111, xyz corb, 10
''')
df = pd.read_csv(data, sep="|")
cols = df.columns.str.split(', ')[0]
s = df.iloc[:,0].str.count(',')
df1 = df.copy()[s.eq(s.min())]
df1 = df1.iloc[:,0].str.split(', ', expand=True)
df1.columns = cols
df2 = df.copy()[s.gt(s.min())]
#inspect this dataframe manually to see how many rows affected, which columns, etc.
#cleanup df2 with some .replace so all equal commas
original = [', inc.', ', corp.']
temp = [' inc.', ' corp.']
df2.iloc[:,0] = df2.iloc[:,0].replace(original, temp, regex=True)
df2 = df2.iloc[:,0].str.split(', ', expand=True)
df2.columns = cols
#cleanup df2 by changing back to original values
df2['col_a'] = df2['col_a'].replace(temp, original, regex=True) # you can do this with other columns as well
df3 = pd.concat([df1, df2]).sort_index()
df3
Out[1]:
col col_a col_b
0 000 abc, inc. 5
1 111 xyz corb 10
Solution 3: Previous Solution (for original question when problem was only in first column - for reference)
You can read in with sep="|" as that | character is not in your .csv, so it reads all of the data into one column.
The main assumption to my solution is that the problematic column is only the first column. I use rsplit(', ') and limit the number of splits to the total number of columns minus 1 (with the example data, this is 2-1=1). Hopefully, this solves with your actual data or at least gives you some idea. If your data is separated by , instead of , , please note whether or not to adjust my splits as well.
import pandas as pd
import numpy as np
from io import StringIO
data = StringIO('''
col_a, col_b
abc, inc., 5
xyz corb, 10
''')
df = pd.read_csv(data, sep="|")
cols = df.columns.str.split(', ')[0]
x = len(cols) - 1
df = df.iloc[:,0].str.rsplit(', ', expand=True, n=x)
df.columns = cols
df
Out[1]:
col_a col_b
0 abc, inc. 5
1 xyz corb 10

convert a list in rows of dataframe in one column to simple string

I have a dataframe which has list in one column that I want to convert into a simple string
id data_words_nostops
26561364 [andrographolide, major, labdane, diterpenoid]
26561979 [dgat, plays, critical, role, hepatic, triglyc]
26562217 [despite, success, imatinib, inhibiting, bcr]
DESIRED OUTPUT
id data_words_nostops
26561364 andrographolide, major, labdane, diterpenoid
26561979 dgat, plays, critical, role, hepatic, triglyc
26562217 despite, success, imatinib, inhibiting, bcr

Try this :
df['data_words_nostops'] = df['data_words_nostops'].apply(lambda row : ','.join(row))
Complete code :
import pandas as pd
l1 = ['26561364', '26561979', '26562217']
l2 = [['andrographolide', 'major', 'labdane', 'diterpenoid'],['dgat', 'plays', 'critical', 'role', 'hepatic', 'triglyc'],['despite', 'success', 'imatinib', 'inhibiting', 'bcr']]
df = pd.DataFrame(list(zip(l1, l2)),
columns =['id', 'data_words_nostops'])
df['data_words_nostops'] = df['data_words_nostops'].apply(lambda row : ','.join(row))
Output :
id data_words_nostops
0 26561364 andrographolide,major,labdane,diterpenoid
1 26561979 dgat,plays,critical,role,hepatic,triglyc
2 26562217 despite,success,imatinib,inhibiting,bcr

df["data_words_nostops"] = df.apply(lambda row: row["data_words_nostops"][0], axis=1)

You can use pandas str join for this:
df["data_words_nostops"] = df["data_words_nostops"].str.join(",")
df
id data_words_nostops
0 26561364 andrographolide,major,labdane,diterpenoid
1 26561979 dgat,plays,critical,role,hepatic,triglyc
2 26562217 despite,success,imatinib,inhibiting,bcr

I tried the following as well
df_ready['data_words_nostops_Joined'] = df_ready.data_words_nostops.apply(', '.join)

How can I create string indexing instead of numbers from dataframe?

I want to create unique row identifiers , in place of the index column, from the contents present in the columns of a dataframe.
For example,
import pandas as pd
from pprint import pprint
df = pd.DataFrame(columns=["ID", "Animal", "Weight", "Description"])
df["ID"] = ["Qw9457", "gft878"]
df["Animal"] = ["Mouse", "Lion"]
df["Weight"] = [20, 67]
df["Description"] = ["hsdg rie", "gtre sjdhi"]
pprint(df)
Output:
ID Animal Weight Description
0 Qw9457 Mouse 20 hsdg rie
1 gft878 Lion 67 gtre sjdhi
I'd prefer to rename the index column using the contents present in the rest of the columns,
for example :
df.index = ["MQwrie", "Lgfgt"]
I would like to know if there are nice ways to programmatically generate
row identifiers(i.e index column) from the contents of columns.

If you are looking to generate an index based on bits of the data in each column, you can piece it together using Series operations and then assign the index. Below, we use the first letter of the animal's name, the weight, and the first word of the description as a new index.
import pandas as pd
df = pd.DataFrame({'ID': ['Qw9457', 'gft878'],
'Animal': ['Mouse', 'Lion'],
'Weight': [20, 67],
'Description': ['hsdg rie', 'gtre sjdhi']})
# create new index from data in df, assign as index
ix = df.Animal.str[0] + df.Weight.astype(str) + df.Description.str.split().str.get(0)
df_new = df.set_index(ix)
df_new
# returns:
ID Animal Weight Description
M20hsdg Qw9457 Mouse 20 hsdg rie
L67gtre gft878 Lion 67 gtre sjdhi
EDIT:
Yes, you add the current row number (starting at zero), you can use:
ix = (
df.Animal.str[0]
+ df.Weight.astype(str)
+ df.Description.str.split().str.get(0)
+ df.index.astype(str).str.zfill(3)
)
df_new = df.set_index(ix)
df_new
#returns:
ID Animal Weight Description
M20hsdg000 Qw9457 Mouse 20 hsdg rie
L67gtre001 gft878 Lion 67 gtre sjdhi

Remove grave accent from IDs

I have an ID column with grave accent like this `1234ABC40 and I want to remove just that character from this column but keep the dataframe form.
I tried this on the column only. I have a file name x here and has multiple columns. id is the col i want to fix.
pd.read_csv(r'C:\filename.csv', index_col = False)
id = str(x['id'])
id2 = unidecode.unidecode(id)
id3 = id2.replace('`','')
This changes to str but I want that column back in the dataframe form

DataFrames have their own replace() function. Note, for partial replacements you must enable regex=True in the parameters:
import pandas as pd
d = {'id': ["12`3", "32`1"], 'id2': ["004`", "9`99"]}
df = pd.DataFrame(data=d)
df["id"] = df["id"].replace('`','', regex=True)
print df
id id2
0 123 004`
1 321 9`99