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Efficient insertion of named tuple into pandas data frame

By : qnm01
Date : September 15 2020, 08:00 AM
With these it helps I have a named tuple called "rows" if I print it, it would look as shown below. How do I efficiently (not in for loop) insert this into a pandas data frame so that I can use it to plot graphs. (Number of rows and columns can go more than a thousand sometimes) , Why not just try:
code :
df = pd.DataFrame(rows)

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Replace all None's in a Pandas data frame with a tuple of None's

By : Tuan Than
Date : March 29 2020, 07:55 AM
wish of those help One way to do it would be using pandas.DataFrame.apply() and pandas.Series.map() like this:
code :
df.apply(lambda ds: ds.map(lambda x: x if x != None else (None, None)))

How to create Pandas data frame from a tuple

By : pyrex.1337
Date : March 29 2020, 07:55 AM
should help you out The answer strongly depends on the String in your tuple. If what you copied is actually whats in the string, you have to convert the string to something pandas can parse, that's why I added the regex substitution.
code :
import pandas as pd
import io
import re
a = (('12','22','32'),
     """Column-1    Column-2    Column-3    colum-4 Column-5    Colum-6 Colum-7 Week    ACCT_YEAR   NAME
12  22  32  …   …   …   …   51  2016    Name-1
12  22  32  …   …   …   …   51  2016    Name-2
12  22  32  …   …   …   …   51  2016    Name-3""")
# The following substitution is only valid if there are absolutely no spaces in values
b = re.sub(string=a[1], pattern=' +', repl=',')
y = pd.read_csv(io.StringIO(b))

Subset pandas data frame based on tuple

By : Dan Baharir
Date : March 29 2020, 07:55 AM
I wish this helpful for you agg+isin
Since tuples are hashable, you can use isin and compare the aggregated values to your last. Using lst and a list directly instead of np.array helps.
code :
>>> lst = [('AA', 'P'), 
           ('BB', 'Q')]

>>> mask = df[['Firstnames', 'Lastnames']].agg(tuple, 1).isin(lst)
>>> df[mask]

    Firstnames  Lastnames   values
0   AA          P           10
1   BB          Q           13
3   AA          P           22
>>> df[mask].sort_values(by=['Firstnames', 'Lastnames'])

    Firstnames  Lastnames   values
0   AA          P           10
3   AA          P           22
1   BB          Q           13
>>> pd.concat([df[df.Firstnames.eq(a) & df.Lastnames.eq(b)] for a,b in lst])

    Firstnames  Lastnames   values
0   AA          P           10
3   AA          P           22
1   BB          Q           13
df = pd.concat([df]*10000).reset_index(drop=True)

%timeit mask = df[['Firstnames', 'Lastnames']].agg(tuple, 1).isin(lst); df[mask].sort_values(by=['Firstnames', 'Lastnames'])
942 ms ± 71.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit pd.concat([df[df.Firstnames.eq(a) & df.Lastnames.eq(b)] for a,b in lst])
16.2 ms ± 355 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
c = list(map(''.join, itertools.product(string.ascii_uppercase, string.ascii_uppercase)))
lst = [(a,b) for a,b in zip(c, list(string.ascii_uppercase)*26)]
df = pd.DataFrame({'Firstnames': c, 'Lastnames': list(string.ascii_uppercase)*26, 'values': 10})

%timeit mask = df[['Firstnames', 'Lastnames']].agg(tuple, 1).isin(lst); df[mask].sort_values(by=['Firstnames', 'Lastnames'])
15.1 ms ± 301 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit pd.concat([df[df.Firstnames.eq(a) & df.Lastnames.eq(b)] for a,b in lst])
781 ms ± 33.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Does pandas read the full data file and stores it in a data frame? Is it efficient to load a 100mb file in pandas?

By : user3015724
Date : March 29 2020, 07:55 AM
I hope this helps you . Personally I am using Pandas for files with sizes between some kilobytes and a few gigabytes without any problems. As stated here Pandas is very efficient with data 100MB up to 1GB. Thats pretty much what I observe when using Pandas.

Python loop through tuple list adding a value from pandas data frame

By : Chelomunoz7
Date : March 29 2020, 07:55 AM
around this issue I am trying to loop through a list of tuples adding a value to the end of each one that corresponds to a value in a column in a pandas data frame. , Use zip, list comprehension solution:
code :
df3 = [j + (i,) for i,j in zip(df2["Beta"], df1)]
for i,j in zip(df2["Beta"], df1):
    j = j + (i,)
[(1, 2, 'a'), (4, 5, 'b'), (8, 9, 'c')] 
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