按名称列表对Pandas中的多个列范围进行切片 [英] Slicing multiple ranges of columns in Pandas, by list of names
问题描述
我正在尝试通过两种不同的方法在Pandas数据框中选择多个列:
I am trying to select multiple columns in a Pandas dataframe in two different approaches:
1)通过列号,例如1-3列和6列起.
1)via the columns number, for examples, columns 1-3 and columns 6 onwards.
和
2)通过列名列表,例如:
2)via a list of column names, for instance:
years = list(range(2000,2017))
months = list(range(1,13))
years_month = list(["A", "B", "B"])
for y in years:
for m in months:
y_m = str(y) + "-" + str(m)
years_month.append(y_m)
然后, years_month 将产生以下内容:
Then, years_month would produce the following:
['A',
'B',
'C',
'2000-1',
'2000-2',
'2000-3',
'2000-4',
'2000-5',
'2000-6',
'2000-7',
'2000-8',
'2000-9',
'2000-10',
'2000-11',
'2000-12',
'2001-1',
'2001-2',
'2001-3',
'2001-4',
'2001-5',
'2001-6',
'2001-7',
'2001-8',
'2001-9',
'2001-10',
'2001-11',
'2001-12']
也就是说,在两种方法中,最好的(或正确的)方式是只加载名称在 years_month 列表中的列?
That said, what is the best(or correct) way to load only the columns in which the names are in the list years_month in the two approaches?
推荐答案
我认为您需要 numpy.r_
合并列的位置,然后使用
I think you need numpy.r_
for concanecate positions of columns, then use iloc
for selecting:
print (df.iloc[:, np.r_[1:3, 6:len(df.columns)]])
以及list
的第二种方法子集:
and for second approach subset by list
:
print (df[years_month])
示例:
df = pd.DataFrame({'2000-1':[1,3,5],
'2000-2':[5,3,6],
'2000-3':[7,8,9],
'2000-4':[1,3,5],
'2000-5':[5,3,6],
'2000-6':[7,8,9],
'2000-7':[1,3,5],
'2000-8':[5,3,6],
'2000-9':[7,4,3],
'A':[1,2,3],
'B':[4,5,6],
'C':[7,8,9]})
print (df)
2000-1 2000-2 2000-3 2000-4 2000-5 2000-6 2000-7 2000-8 2000-9 A \
0 1 5 7 1 5 7 1 5 7 1
1 3 3 8 3 3 8 3 3 4 2
2 5 6 9 5 6 9 5 6 3 3
B C
0 4 7
1 5 8
2 6 9
print (df.iloc[:, np.r_[1:3, 6:len(df.columns)]])
2000-2 2000-3 2000-7 2000-8 2000-9 A B C
0 5 7 1 5 7 1 4 7
1 3 8 3 3 4 2 5 8
2 6 9 5 6 3 3 6 9
您还可以将ranges
的总和(必须在python 3
中广播到list
):
You can also sum of ranges
(cast to list
in python 3
is necessary):
rng = list(range(1,3)) + list(range(6, len(df.columns)))
print (rng)
[1, 2, 6, 7, 8, 9, 10, 11]
print (df.iloc[:, rng])
2000-2 2000-3 2000-7 2000-8 2000-9 A B C
0 5 7 1 5 7 1 4 7
1 3 8 3 3 4 2 5 8
2 6 9 5 6 3 3 6 9
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