Pandas：按索引计算细胞频率

Question

我的数据框是一长串 4 个字母'A', 'T', 'G','C'，我需要按索引计算每个字母的频率

df = pd.DataFrame({'cases': ['ACCTTGTAGTGTATTTTATGACCAAATGACTTTTTCCCCCCAGTGGCTAATTTGTCTCAGGCCTGCGTCTTAAAGAGACACGGTAATGAGTAGGAAGTCCAGCGTGGTCTGGA','ACCTTGTACTGTATCTTATGACCAGATGACTTTTTCCACCCAGTGGCTAATTTGTCTCAGGCCTCCGTCTTAAAGAGACACGGTAATGAGTAGGAAGTCCAACGTGGTCTAGA','GCCTTGTACTGTATATTATGACCAAATGACTTTTTCCACCCATTGGCTAATTTGTCTCAGGCCTCCGTCTTAAAGAGACACGGAAATGAGTAGGAAGTCCAGCGTGGTCTAGA','ACCTTGTACTGTATATTATGACCAGATGACTTTTTCCACCCAGTGGCTAATTTGTCTCAGGCCTCCGTCTTAAAGAGACACGGTAATGAGTAGGAAGTCCAGCGTGGTCTAGA']})

                                               cases
0  ACCTTGTAGTGTATTTTATGACCAAATGACTTTTTCCCCCCAGTGG...
1  ACCTTGTACTGTATCTTATGACCAGATGACTTTTTCCACCCAGTGG...
2  GCCTTGTACTGTATATTATGACCAAATGACTTTTTCCACCCATTGG...
3  ACCTTGTACTGTATATTATGACCAGATGACTTTTTCCACCCAGTGG...
4  ACCTTGTACTGTATATTATGACCAGATGACTTTTTCCACCCAGTGG...
5  ACCTTGTAGTGTATTTTATGACCAAATGACTTTTTCCCCCCAGTGG...
6  ACCTTGTACTGTATCTTATGACCAGATGACTTTTTCCACCCAGTGG...
7  GCCTTGTACTGTATATTATGACCAAATGACTTTTTCCACCCATTGG...
8  ACCTTGTACTGTATATTATGACCAGATGACTTTTTCCACCCAGTGG...
9  ACCTTGTACTGTATATTATGACCAGATGACTTTTTCCACCCAGTGG...

结果将是一个新的 df 形状4x113，我想不出一个熊猫的方法来做到这一点。以下是我的非熊猫解决方案

def freq_lists(dna_list):
    n = len(dna_list[0])
    A = [0]*n
    T = [0]*n
    G = [0]*n
    C = [0]*n
    for dna in dna_list:
        for index, base in enumerate(dna):
            if base == 'A':
                        A[index] += 1
            elif base == 'C':
                C[index] += 1
            elif base == 'G':
                        G[index] += 1
            elif base == 'T':
                T[index] += 1
    return {'A': A, 'C': C, 'G': G, 'T': T}

fdf = pd.DataFrame(freq_lists(df['cases'].to_list()))

     A  C  G  T
0    3  0  1  0
1    0  4  0  0
2    0  4  0  0
3    0  0  0  4
4    0  0  0  4
..  .. .. .. ..
108  0  4  0  0
109  0  0  0  4
110  3  0  1  0
111  0  0  4  0
112  4  0  0  0

为了澄清第一行是通过总结case列中第一个 str 的计数获得的AAGA -> A: 3, C:0, G:1 T:0

Answer 1

让我们做explode与crosstab

s = df.cases.map(list).explode()
out = pd.crosstab(s.groupby(level=0).cumcount(),s)
Out[583]: 
cases  A  C  G  T
row_0            
0      3  0  1  0
1      0  4  0  0
2      0  4  0  0
3      0  0  0  4
4      0  0  0  4
   .. .. .. ..
108    0  4  0  0
109    0  0  0  4
110    3  0  1  0
111    0  0  4  0
112    4  0  0  0

Answer 2

使用集合。计数器：

from collections import Counter
df['cases'].apply(lambda x: pd.Series(Counter(x)))

输出：

    A   C   T   G
0  27  24  34  28
1  29  26  33  25
2  30  25  33  25
3  29  25  33  26

反之则不那么性感：

pd.DataFrame([Counter(i)
              for i in list(zip(*df['cases'].apply(list).values))]
            ).fillna(0).astype(int)

或者

(df['cases'].apply(lambda x: pd.Series(list(x)))
            .apply(pd.value_counts)
            .T.fillna(0).astype(int)
)

输出：

     A  G  C  T
0    3  1  0  0
1    0  0  4  0
2    0  0  4  0
...
111  0  4  0  0
112  4  0  0  0