Creating and overwriting dataframe(lists) in a loop in Python

Question

This is a powerful simplification of my earlier problem :)

There are 4 lists (In my case the values are integers):

e00 = [AB, CA]
e01 = [BD, FK]
e10 = [AC, DF]
e11 = [DE, FB]

a = input('Write a binary number. \n')

    b = []
    for i in range(len(a)-1):
        c = a[i:i+2]
        b.append(c)

For example, for string 00101, I get:

b = ['00', '01', '10', '01']

Stages:

I choose the first two values in list 'b', ie '00' and '01'
For this reason, I will use lists e00 and e01

I compare the above lists. My code just for this case:

a00 = pd.DataFrame({'A':e00})
a01 = pd.DataFrame({'B':e01})
a10 = pd.DataFrame({'C':e10})
a11 = pd.DataFrame({'D':e11})

    b = []
    for i in range(len(a)-1):
        c = a[i:i+2]
        b.append(c)

if b[0]=='00':
   if b[1]=='01':

       m001 = a00.assign(x=a00.A.str[-1]).merge(a01.assign(x=a01.B.str[0], B=a01.B.str[1:]))

       v = (m001['A'] + m001['B'])
       v.str.len() == v.apply(set).str.len()
       c=v[v.str.len() == v.apply(set).str.len()].tolist()

Which gives:

c = [ABD]

I choose the third value from the list 'b', ie '10'
For this reason I will use c and e10
I compare the above lists.
I get c = [ABDF]
I choose the last fourth value in the list 'b', ie '01'
For this reason I will use c and e01
I compare the above lists.
Finally, I get c = [ABDFK]
End

I hope now the problem will be brighter :)

minimal reproducible example if you can? Along with expected output, if you can reduce the explanation. Thanks! — cs95
– cs95, Commented Nov 19, 2017 at 18:40
@ cᴏʟᴅsᴘᴇᴇᴅ This fragment above works alone. just yet of course import pandas as pd :) — Tomasz Przemski
– Tomasz Przemski, Commented Nov 19, 2017 at 18:47
I believe coldspeed is asking for more readability, in particular, shorten the explanation and break into 2 or 3 smaller paragraphs instead of just one big one. And often, just showing a simple example of desired output is easier to understand than a long explanation. — JohnE
– JohnE, Commented Nov 19, 2017 at 18:50

cs95 · Accepted Answer · 2017-11-19 20:26:01Z

2

If I understood you clearly, you're looking for a way to dynamically access aXX so you can then use them to perform your computation.

What you need is a dictionary.

dct = {'00' : a00, '01' : a01, '10' : a10, '11' : a11}    
c = dct[b[0]]

for i in b[1:]:
    d = dct[i]
    newcols = {'x' : d.iloc[:, 0].str[0], d.columns[0] : d.iloc[:, 0].str[1:]}
    m = c.assign(x=c.iloc[:, 0].str[-1]).merge(d.assign(**newcols))

    ... # rest of your code

You can use b's elements to index into dct and extract the dataframes you need for that iteration.

edited Nov 19, 2017 at 20:26

answered Nov 19, 2017 at 20:01

cs95

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4 Comments

Tomasz Przemski Over a year ago

you are great :) Everything works great on the above case. But I'm trying to modify it now for all frames a00, a01, a10, a11, because in each frame I fixed a different column name. And in 'm' only A and B are used.

cs95 Over a year ago

@TomaszPrzemski You can refer to the first column with df.iloc[:, 0].

cs95 Over a year ago

@TomaszPrzemski I see the problem of dynamic columns is a bigger issue than I thought. So I'm initialising a dictionary of new columns and passing that to d after. It's easier. Take a look at my latest edit, I hope it helps.

Tomasz Przemski Over a year ago

However, I will still have to sit down for the night :) Since probably 'dict' blocks certain cases like "000". And the question of using longer binary strings is open. For now it works only on three digits :) But and anyway you helped me so much!

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