Find the difference between strings for each two rows of pandas data.frame

Question

I am new in python, and I am struggling with this for some time. I have a file that looks like this:

    name   seq
1   a1     bbb
2   a2     bbc
3   b1     fff
4   b2     fff
5   c1     aaa
6   c2     acg

where name is the name of the string and seq is the string. I would like a new column or a new data frame that indicates the number of differences between every two rows without overlap. For example, I want the number of differences between sequences for the name [a1-a2] then [b1-b2] and lastly between [c1-c2].

So I need something like this:

    name   seq   diff  
1   a1     bbb    NA   
2   a2     bbc    1
3   b1     fff    NA
4   b2     fff    0
5   c1     aaa    NA
6   c2     acg    2

Any help is highly appreciated

Answer 1

It looks like you want the jaccard distance of the pairs of strings. Here's one way using groupby and scipy.spatial.distance.jaccard:

from scipy.spatial.distance import jaccard
g = df.groupby(df.name.str[0])

df['diff'] = [sim for _, seqs in g.seq for sim in 
              [float('nan'), jaccard(*map(list,seqs))]]

---

print(df)

  name  seq  diff
1   a1  bbb   NaN
2   a2  bbc   1.0
3   b1  fff   NaN
4   b2  fff   0.0
5   c1  aaa   NaN
6   c2  acg   2.0

Answer 2

Alternative with Levenshtein distance:

import Levenshtein
s = df['name'].str[0]
out = df.assign(Diff=s.drop_duplicates(keep='last').map(df.groupby(s)['seq']
                    .apply(lambda x: Levenshtein.distance(x.iloc[0],x.iloc[-1]))))

---

  name  seq  Diff
1   a1  bbb   NaN
2   a2  bbc   1.0
3   b1  fff   NaN
4   b2  fff   0.0
5   c1  aaa   NaN
6   c2  acg   2.0

Answer 3

As a first step I recreated your data with:

#!/usr/bin/env python3
import pandas as pd

# Setup
data = {'name': {1: 'a1', 2: 'a2', 3: 'b1', 4: 'b2', 5: 'c1', 6: 'c2'}, 'seq': {1: 'bbb', 2: 'bbc', 3: 'fff', 4: 'fff', 5: 'aaa', 6: 'acg'}}
df = pd.DataFrame(data)

Solution You could try to iterate over the dataframe and compare the seq value of the last iteration with the current one. For the comparison of the two strings (stored in the seq columns of your dataframe) you can apply a simple list comprehension like in this function:

def diff_letters(a,b):
    return sum ( a[i] != b[i] for i in range(len(a)) )

Iteration over the Dataframe rows

diff = ['NA']

row_iterator = df.iterrows()
_, last = next(row_iterator)

# Iterate over the df get populate a list with result of the comparison
for i, row in row_iterator:
    if i % 2 == 0:
        diff.append(diff_letters(last['seq'],row['seq']))
    else:
        # for odd row numbers append NA value
        diff.append("NA")
    last = row
df['diff'] = diff

Result looks like this

  name  seq diff
1   a1  bbb   NA
2   a2  bbc    1
3   b1  fff   NA
4   b2  fff    0
5   c1  aaa   NA
6   c2  acg    2

Answer 4

Check this one

import pandas as pd

data = {'name':  ['a1', 'a2','b1','b2','c1','c2'],
    'seq': ['bbb', 'bbc','fff','fff','aaa','acg']
    }

df = pd.DataFrame (data, columns = ['name','seq'])
diffCntr=0
df['diff'] = np.nan
i=0
while i < len(df)-1:
    diffCntr=np.nan
    item=df.at[i,'seq']
    df.at[i,'diff']=diffCntr
    diffCntr=0
    for j in df.at[i+1,'seq']:
        if item.find(j) < 0:
            diffCntr +=1
    df.at[i+1,'diff']=diffCntr
    i +=2    
df  

The result is this:

    name seq    diff
0   a1   bbb    NaN
1   a2   bbc    1.0
2   b1   fff    NaN
3   b2   fff    0.0
4   c1   aaa    NaN
5   c2   acg    2.0