Finding the count of letters in each column

Question

I need to find the count of letters in each column as follows:

String: ATCG
        TGCA
        AAGC
        GCAT

string is a series.

I need to write a program to get the following:

  0 1 2 3
A 2 1 1 1
T 1 1 0 1
C 0 1 2 1
G 1 1 1 1 

I have written the following code but I am getting a row in 0 index and column at the end (column index 450, actual column no 451) with nan values. I should not be getting either the row or the column 451. I need to have only 450 columns.

f = zip(*string)
counts = [{letter: column.count(letter) for letter in column} for column in 
f]
counts=pd.DataFrame(counts).transpose()
print(counts)
counts = counts.drop(counts.columns[[450]], axis =1)

Can anyone please help me understand the issue?

Answer 1

Here is one way you can implement your logic. If required, you can turn your series into a list via lst = s.tolist() .

lst = ['ATCG', 'TGCA', 'AAGC', 'GCAT']

arr = [[i.count(x) for i in zip(*lst)] for x in ('ATCG')]

res = pd.DataFrame(arr, index=list('ATCG'))

Result

   0  1  2  3
A  2  1  1  1
T  1  1  0  1
C  0  1  2  1
G  1  1  1  1

Explanation

• In the list comprehension, deal with columns first by iterating the first, second, third and fourth elements of each string sequentially.
• Deal with rows second by iterating through 'ATCG' sequentially.
• This produces a list of lists which can be fed directly into pd.DataFrame .

Answer 2

With Series.value_counts() :

>>> s = pd.Series(['ATCG', 'TGCA', 'AAGC', 'GCAT'])

>>> s.str.join('|').str.split('|', expand=True)\
...     .apply(lambda row: row.value_counts(), axis=0)\
...     .fillna(0.)\
...     .astype(int)
   0  1  2  3
A  2  1  1  1
C  0  1  2  1
G  1  1  1  1
T  1  1  0  1

I'm not sure how logically you want to order the index, but you could call .reindex() or .sort_index() on this result.

The first line, s.str.join('|').str.split('|', expand=True) gets you an "expanded" version

   0  1  2  3
0  A  T  C  G
1  T  G  C  A
2  A  A  G  C
3  G  C  A  T

which should be faster than calling pd.Series(list(x)) ... on each row.