CODEWITHSUNDEEP
pythonpandas
user3329732
user3329732

Reputation: 343

How can I convert columns based on multiple contingencies of other columns?

I have a dataframe:

df = pd.DataFrame({'REF':list('GCTT'), 'ALT':list('AACG'),
                   'A1':['0/1','0/1','0/0','0/1'],
                   'A2':['1/1','0/1','0/1','0/0']})

  REF ALT   A1   A2
0   G   A  0/1  1/1
1   C   A  0/1  0/1
2   T   C  0/0  0/1
3   T   G  0/1  0/0

I want to convert the columns A1 and A2 based on values in the REF and ALT columns. So, columns A1 & A2 in Row 0 should read GA & AA. ie lose the "/" and replace 0s with G and 1s with A. Next, row 1 should replace 0s with C and 1s with A. Then follow the pattern for the next rows, to get:

  REF ALT  A1   A2
0   G   A  GA  AA
1   C   A  CA  CA
2   T   C  TT  TC
3   T   G  TG  TT

In my data there are hundreds of A columns: A1, A2......An-1, An. So, the solution needs to be replicable across all columns.

Upvotes: 2

Views: 57

Answers (3)

kantal
kantal

Reputation: 2407

I wonder how fast this solution is with your data:

for col in ["A1","A2"]: 
        df[col]= df[col].str.split("/",expand=True) \
                        .replace(["0","1"],[df.REF,df.ALT]) \
                        .agg("".join,axis=1) 

df                                                                                                                  

  REF ALT  A1  A2
0   G   A  GA  AA
1   C   A  CA  CA
2   T   C  TT  TC
3   T   G  TG  TT

EDIT: Solution 2., Working with indexes:

# helper structs:
ncbscols= ["REF","ALT"]
cols= df.columns.difference(ncbscols)

ii= pd.MultiIndex.from_product([list("ACGT"),list("ACGT"),["0/0","0/1","1/1","1/0"] ])
ser= pd.Series( [t[2].replace("/","").replace("0",t[0]).replace("1",t[1]) for t in ii ],  index=ii )

# the main calculation:
for c in cols:
    mi= pd.MultiIndex.from_arrays([ df.REF.values,df.ALT.values,df[c].values ])
    df[c]= ser[mi].values


ser:
 A  A  0/0    AA
      0/1    AA
      1/1    AA
      1/0    AA
   C  0/0    AA
             ..
T  G  1/0    GT
   T  0/0    TT
      0/1    TT
      1/1    TT
      1/0    TT
Length: 64, dtype: object 

df:
  REF ALT  A1  A2
0   G   A  GA  AA
1   C   A  CA  CA
2   T   C  TT  TC
3   T   G  TG  TT

Upvotes: 2

Andy L.
Andy L.

Reputation: 25239

You have only 4 combo of 0 and 1, so I think you may try np.select

df1 = df.drop(['REF', 'ALT'], axis=1)

#conditions
combo = ['0/0', '0/1', '1/0', '1/1']
conds = [df1.eq(x) for x in combo]

#selections
s00 = (df.REF * 2).to_numpy()[:,None]
s11 = (df.ALT * 2).to_numpy()[:,None]
s01 = (df.REF + df.ALT).to_numpy()[:,None]

df.loc[:, df1.columns.tolist()] = np.select(conds , [s00, s01, s01[:,::-1], s11], np.nan)

Out[260]:
  REF ALT  A1  A2
0   G   A  GA  AA
1   C   A  CA  CA
2   T   C  TT  TC
3   T   G  TG  TT

Upvotes: 0

Kenan
Kenan

Reputation: 14094

I think there might be a better way to do this, but this works. let me know if it's a performance issue.

acols = df.drop(['REF', 'ALT'], axis=1).columns

for i in acols:
    df.loc[df[i] == '0/0', i] = df['REF'] * 2
    df.loc[df[i] == '0/1', i] = df['REF'] + df['ALT']
    df.loc[df[i] == '1/1', i] = df['ALT'] * 2

Another option

for i in acols:
    df[i] = df[i].replace(to_replace='0/0', value=df['REF']+df['REF'])
    df[i] = df[i].replace(to_replace='0/1', value=df['REF']+df['ALT'])
    df[i] = df[i].replace(to_replace='1/1', value=df['ALT']+df['ALT'])

  REF ALT  A1  A2
0   G   A  GA  AA
1   C   A  CA  CA
2   T   C  TT  TC
3   T   G  TG  TT

Upvotes: 0

Related Questions