我需要从10个列表中构建一个数据帧。我是手工做的,但需要一段时间。有什么更好的方法?你知道吗
我试过手工做。工作正常(#1) 我尝试了代码(#2)以获得更好的性能,但它只返回最后一列。你知道吗
import pandas as pd
import numpy as np
a1T=[([7,8,9]),([10,11,12]),([13,14,15])]
a2T=[([1,2,3]),([5,0,2]),([3,4,5])]
print (a1T)
#Output[[7, 8, 9], [10, 11, 12], [13, 14, 15]]
vis1=np.array (a1T)
vis_1_1=vis1.T
tmp2=np.array (a2T)
tmp_2_1=tmp2.T
X=np.column_stack([vis_1_1, tmp_2_1])
dataset_all = pd.DataFrame({"Visab1":X[:,0], "Visab2":X[:,1], "Visab3":X[:,2], "Temp1":X[:,3], "Temp2":X[:,4], "Temp3":X[:,5]})
print (dataset_all)
Output: Visab1 Visab2 Visab3 Temp1 Temp2 Temp3
0 7 10 13 1 5 3
1 8 11 14 2 0 4
2 9 12 15 3 2 5
> Actually I have varying number of columns in dataframe (500-1500), thats why I need auto generated column names. Extra index (1, 2, 3) after name Visab_, Temp_ and so on - constant for every case. See code below.
For better perfomance I tried
code<br>
#2
n=3 # This is varying parameter. The parameter affects the number of columns in the table.
m=2 # This is constant for every case. here is 2, because we have "Visab", "Temp"
mlist=('Visab', 'Temp')
nlist=[range(1, n)]
for j in range (1,n):
for i in range (1,m):
col=i+(j-1)*n
dataset_all=pd.DataFrame({mlist[j]+str(i):X[:, col]})
I expect output like
Visab1 Visab2 Visab3 Temp1 Temp2 Temp3
0 7 10 13 1 5 3
1 8 11 14 2 0 4
2 9 12 15 3 2 5
but there is not any result (only error expected an indented block)
现在更清楚了。所以你有:
让我们创建一个列名称列表:
现在您可以直接创建如下数据帧:
好的,那么n列的数量就是每个列表中的子列表的数量,对吗?你可以用len测量:
我将简化上面的答案,这样您就不需要使用X并添加自动列名创建:
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