from io import StringIO
import pandas as pd
import numpy as np
txt = """pos M1 M2 M3 M4 M5 M6 M7 M8 hybrid_block S1 S2 S3 S4 S5 S6 S7 S8
1 A T T A A G A C A|C C G C T T A G A
2 T G C T G T T G T|A A T A T C A A T
3 C A A C A G T C C|G G A C G C G C G
4 G T G T A T C T G|T C T T T A T C T """
df = pd.read_csv(StringIO(txt), delim_whitespace=True, index_col='pos')
df
d1 = pd.concat([df.loc[[1]].rename(index={1: 0}), df])
d1 = pd.concat([
df.filter(like='M'),
df.hybrid_block.str.split('|', expand=True).rename(columns='H{}'.format),
df.filter(like='S')
], axis=1)
d1 = pd.concat([d1.loc[[1]].rename(index={1: 0}), d1])
d1 = d1.add('g').add(d1.shift()).dropna()
d1
将方便的块分配给自己的变量名
m = d1.filter(like='M')
s = d1.filter(like='S')
h = d1.filter(like='H')
计算每个块中有多少并连接
mcounts = pd.DataFrame(
(m.values[:, :, None] == h.values[:, None, :]).sum(1),
h.index, h.columns
)
scounts = pd.DataFrame(
(s.values[:, :, None] == h.values[:, None, :]).sum(1),
h.index, h.columns
)
counts = pd.concat([mcounts, scounts], axis=1, keys=['M', 'S'])
counts
如果你真的想要字典
d = defaultdict(lambda:defaultdict(list))
dict_df = counts.stack().join(h.stack().rename('condition')).unstack()
for pos, row in dict_df.iterrows():
d['M']['H0'].append((row.loc[('condition', 'H0')], row.loc[('M', 'H0')]))
d['S']['H0'].append((row.loc[('condition', 'H0')], row.loc[('S', 'H0')]))
d['M']['H1'].append((row.loc[('condition', 'H1')], row.loc[('M', 'H1')]))
d['S']['H1'].append((row.loc[('condition', 'H1')], row.loc[('S', 'H1')]))
dict(d)
{'M': defaultdict(list,
{'H0': [('AgA', 4), ('TgA', 3), ('CgT', 2), ('GgC', 1)],
'H1': [('CgC', 1), ('AgC', 0), ('GgA', 0), ('TgG', 1)]}),
'S': defaultdict(list,
{'H0': [('AgA', 2), ('TgA', 1), ('CgT', 0), ('GgC', 0)],
'H1': [('CgC', 2), ('AgC', 2), ('GgA', 2), ('TgG', 3)]})}
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