Question

给出两个数组：

x
[('010_628', '2543677'), ('010_228', '2543677'), ('015_634', '2543677')]

y 
array([['me', 10228955],
       ['me', 10228955],
       ['me', 10228955]], dtype=object)

目前，这段代码为我提供了一个平面索引为元组的数据框：

df = pd.DataFrame(x, index=y, columns=['pm_code',   'sec_pm'])
df
                pm_code   sec_pm
(me, 10228955)  010_628  2543677
(me, 10228955)  010_228  2543677
(me, 10228955)  015_634  2543677

如何创建一个看起来像这样的MultiIndex数据框？

                  pm_code   sec_pm
state site_no                     
me    10228955   010_628  2543677
                 010_228  2543677
                 015_634  2543677

我尝试使用pd.MultiIndex.from_tuples，但我无法做到这一点。谢谢你的帮助。

附录：效果比较

小

# unutbu #1
%timeit pd.DataFrame(x, index=pd.MultiIndex.from_arrays(y.T), columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.25 ms per loop

# unutbu #2
%timeit pd.DataFrame(x, index=pd.MultiIndex.from_tuples(y.tolist()), columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.47 ms per loop

# piRSquared
%timeit pd.DataFrame(x, index=y.T.tolist(), columns=['pm_code', 'sec_pm'])
1000 loops, best of 3: 1.41 ms per loop

# Andrew L
%timeit pd.DataFrame(x, index=[y[:,0], y[:,1]], columns=['pm_code',   'sec_pm'])
1000 loops, best of 3: 1.29 ms per loop

大

x2 = np.repeat(x, 10000, 0)
y2 = np.repeat(x, 10000, 0)

# unutbu #1
%timeit pd.DataFrame(x2, index=pd.MultiIndex.from_arrays(y2.T), columns=['pm_code',   'sec_pm'])
100 loops, best of 3: 17.3 ms per loop

# unutbu #2
%timeit pd.DataFrame(x2, index=pd.MultiIndex.from_tuples(y2.tolist()), columns=['pm_code',   'sec_pm'])
10 loops, best of 3: 30.5 ms per loop

# piRSquared
%timeit pd.DataFrame(x2, index=y2.T.tolist(), columns=['pm_code', 'sec_pm'])
10 loops, best of 3: 37.2 ms per loop

# Andrew L
%timeit pd.DataFrame(x2, index=[y2[:,0], y2[:,1]], columns=['pm_code',   'sec_pm'])
100 loops, best of 3: 22 ms per loop

_{此question的数据。}

Answer 1

您可以使用pd.MultiIndex.from_arrays(y.T)：

In [53]: pd.DataFrame(x, index=pd.MultiIndex.from_arrays(y.T), columns=['pm_code',   'sec_pm'])
Out[53]: 
             pm_code   sec_pm
me 10228955  010_628  2543677
   10228955  010_228  2543677
   10228955  015_634  2543677

或pd.MultiIndex.from_tuples(y.tolist())：

In [54]: pd.DataFrame(x, index=pd.MultiIndex.from_tuples(y.tolist()), columns=['pm_code',   'sec_pm'])
Out[54]: 
             pm_code   sec_pm
me 10228955  010_628  2543677
   10228955  010_228  2543677
   10228955  015_634  2543677

Answer 2

您还可以对数组进行切片并传递给awk -F '|' '"ABC" $2 "DEF" == $1 && "ABC" $2 "DEFZ"+[0-9] == $1 { print }' WHTFile.txt > QC2Valid.txt**：

index

Answer 3

选项1
如果你传递一个类似于数组的列表，那么构造函数就知道如何处理它。

pd.DataFrame(x, index=y.T.tolist(), columns=['pm_code', 'sec_pm'])

    pm_code   sec_pm
me 10228955  010_628  2543677
   10228955  010_228  2543677
   10228955  015_634  2543677

通过构造函数创建MultiIndexed数据帧

小

大

3 个答案: