我认为您需要在末尾添加 ,: - 这意味着您需要切片行,但需要所有列:
print (df.loc[('chr1', slice(3000714, 3001110)),:])
end ref|alt
chrom start
chr1 3000714 3000715 T|G
3001065 3001066 G|T
3001110 3001111 G|C
另一种解决方案是将axis=0添加到loc:
print (df.loc(axis=0)[('chr1', slice(3000714, 3001110))])
end ref|alt
chrom start
chr1 3000714 3000715 T|G
3001065 3001066 G|T
3001110 3001111 G|C
但如果只需要3000714 和3001110:
print (df.loc[('chr1', [3000714, 3001110]),:])
end ref|alt
chrom start
chr1 3000714 3000715 T|G
3001110 3001111 G|C
idx = pd.IndexSlice
print (df.loc[idx['chr1', [3000714, 3001110]],:])
end ref|alt
chrom start
chr1 3000714 3000715 T|G
3001110 3001111 G|C
时间安排:
In [21]: %timeit (df.loc[('chr1', slice(3000714, 3001110)),:])
1000 loops, best of 3: 757 µs per loop
In [22]: %timeit (df.loc(axis=0)[('chr1', slice(3000714, 3001110))])
1000 loops, best of 3: 743 µs per loop
In [23]: %timeit (df.loc[('chr1', [3000714, 3001110]),:])
1000 loops, best of 3: 824 µs per loop
In [24]: %timeit (df.loc[pd.IndexSlice['chr1', [3000714, 3001110]],:])
The slowest run took 5.35 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 826 µs per loop