<p>下面是一种方法,使用<code>Counter</code>:</p>
<pre><code>from collections import Counter
data = {'chr2:173370685-173370692': 'TACCAAG', 'chr5:118309829-118309836': 'TCTCCTT', 'chr12:104659651-104659658': 'GACCAAA'}
out = {i:Counter(col) for i, col in enumerate(zip(*(data.values()))) }
# we can add the missing keys whose count is 0:
for count in out.values():
count.update(dict.fromkeys('ATGC', 0))
print(out)
# {0: Counter({'T': 2, 'G': 1, 'A': 0, 'C': 0}), 1: Counter({'A': 2, 'C': 1, 'T': 0, 'G': 0}),
# 2: Counter({'C': 2, 'T': 1, 'A': 0, 'G': 0}), 3: Counter({'C': 3, 'A': 0, 'T': 0, 'G': 0}),
# 4: Counter({'A': 2, 'C': 1, 'T': 0, 'G': 0}), 5: Counter({'A': 2, 'T': 1, 'G': 0, 'C': 0}),
# 6: Counter({'G': 1, 'T': 1, 'A': 1, 'C': 0})}
</code></pre>
<p>我把原始索引保留为整数,使用它们可能比使用像“1”、“2”这样的字符串更容易。。。但如果你真的想:</p>
<pre><code>numbers_as_strings = ['one', 'two', 'three', 'four', 'five', 'six', 'seven']
out = {numbers_as_strings[key]:value for key, value in out.items()}
print(out)
# {'one': Counter({'T': 2, 'G': 1, 'A': 0, 'C': 0}),
# 'two': Counter({'A': 2, 'C': 1, 'T': 0, 'G': 0}) ....
</code></pre>