Python 计算熊猫数据框中选定列的选定行的平均值
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Calculate mean for selected rows for selected columns in pandas data frame
提问by impossible
I have pandas df with say, 100 rows, 10 columns, (actual data is huge). I also have row_index list which contains, which rows to be considered to take mean. I want to calculate mean on say columns 2,5,6,7 and 8. Can we do it with some function for dataframe object?
我有熊猫 df 说，100 行，10 列，（实际数据很大）。我也有 row_index 列表，其中包含哪些行被认为是平均的。我想计算第 2、5、6、7 和 8 列的平均值。我们可以用一些数据框对象的函数来做吗？
What I know is do a for loop, get value of row for each element in row_index and keep doing mean. Do we have some direct function where we can pass row_list, and column_list and axis, for ex df.meanAdvance(row_list,column_list,axis=0)?
我所知道的是做一个 for 循环，获取 row_index 中每个元素的行值并保持平均。我们是否有一些直接的函数，我们可以在其中传递 row_list、column_list 和轴，例如df.meanAdvance(row_list,column_list,axis=0)？
I have seen DataFrame.mean() but it didn't help I guess.
我见过 DataFrame.mean() 但我猜它没有帮助。
a b c d q
0 1 2 3 0 5
1 1 2 3 4 5
2 1 1 1 6 1
3 1 0 0 0 0
I want mean of 0, 2, 3rows for each a, b, dcolumns
我想要0, 2, 3每a, b, d列的行数
a b d
0 1 1 2
采纳答案by PdevG
To select the rows of your dataframe you can use iloc, you can then select the columns you want using square brackets.
要选择数据框的行，您可以使用 iloc，然后您可以使用方括号选择所需的列。
For example:
例如：
df = pd.DataFrame(data=[[1,2,3]]*5, index=range(3, 8), columns = ['a','b','c'])
gives the following dataframe:
给出以下数据框：
a b c
3 1 2 3
4 1 2 3
5 1 2 3
6 1 2 3
7 1 2 3
to select only the 3d and fifth row you can do:
要仅选择 3d 和第五行，您可以执行以下操作：
df.iloc[[2,4]]
which returns:
返回：
a b c
5 1 2 3
7 1 2 3
if you then want to select only columns b and c you use the following command:
如果您只想选择列 b 和 c，则使用以下命令：
df[['b', 'c']].iloc[[2,4]]
which yields:
产生：
b c
5 2 3
7 2 3
To then get the mean of this subset of your dataframe you can use the df.mean function. If you want the means of the columns you can specify axis=0, if you want the means of the rows you can specify axis=1
然后，您可以使用 df.mean 函数来获得数据帧的这个子集的平均值。如果你想要列的平均值，你可以指定axis=0，如果你想要行的平均值，你可以指定axis=1
thus:
因此：
df[['b', 'c']].iloc[[2,4]].mean(axis=0)
returns:
返回：
b 2
c 3
As we should expect from the input dataframe.
正如我们对输入数据帧所期望的那样。
For your code you can then do:
对于您的代码，您可以执行以下操作：
df[column_list].iloc[row_index_list].mean(axis=0)
EDIT after comment: New question in comment: I have to store these means in another df/matrix. I have L1, L2, L3, L4...LX lists which tells me the index whose mean I need for columns C[1, 2, 3]. For ex: L1 = [0, 2, 3] , means I need mean of rows 0,2,3 and store it in 1st row of a new df/matrix. Then L2 = [1,4] for which again I will calculate mean and store it in 2nd row of the new df/matrix. Similarly till LX, I want the new df to have X rows and len(C) columns. Columns for L1..LX will remain same. Could you help me with this?
评论后编辑：评论中的新问题：我必须将这些方法存储在另一个 df/matrix 中。我有 L1、L2、L3、L4...LX 列表，它告诉我我需要列 C[1,2,3] 的平均值的索引。例如： L1 = [0, 2, 3] ，意味着我需要行 0,2,3 的平均值并将其存储在新 df/matrix 的第一行中。然后 L2 = [1,4] 为此我将再次计算平均值并将其存储在新 df/matrix 的第二行中。同样，直到 LX，我希望新的 df 具有 X 行和 len(C) 列。L1..LX 的列将保持不变。你能帮我解决这个问题吗？
Answer:
回答：
If i understand correctly, the following code should do the trick (Same df as above, as columns I took 'a' and 'b':
如果我理解正确，下面的代码应该可以解决问题（与上面的 df 相同，因为我采用了 'a' 和 'b' 列：
first you loop over all the lists of rows, collection all the means as pd.series, then you concatenate the resulting list of series over axis=1, followed by taking the transpose to get it in the right format.
首先循环遍历所有行列表，将所有均值收集为 pd.series，然后在轴 = 1 上连接结果序列列表，然后进行转置以获得正确的格式。
dfs = list()
for l in L:
dfs.append(df[['a', 'b']].iloc[l].mean(axis=0))
mean_matrix = pd.concat(dfs, axis=1).T
回答by mfitzp
You can select specific columns from a DataFrame by passing a list of indices to .iloc, for example:
您可以通过将索引列表传递给.iloc，从 DataFrame 中选择特定列，例如：
df.iloc[:, [2,5,6,7,8]]
Will return a DataFrame containing those numbered columns (note: This uses 0-based indexing, so 2refers to the 3rd column.)
将返回一个包含这些编号列的 DataFrame（注意：这使用基于 0 的索引，因此2指的是第 3 列。）
To take a mean down of that column, you could use:
要降低该列的平均值，您可以使用：
# Mean along 0 (vertical) axis: return mean for specified columns, calculated across all rows
df.iloc[:, [2,5,6,7,8]].mean(axis=0)
To take a mean across that column, you could use:
要在该列中取平均值，您可以使用：
# Mean along 1 (horizontal) axis: return mean for each row, calculated across specified columns
df.iloc[:, [2,5,6,7,8]].mean(axis=1)
You can also supply specific indices for both axes to return a subset of the table:
您还可以为两个轴提供特定索引以返回表的子集：
df.iloc[[1,2,3,4], [2,5,6,7,8]]
For your specific example, you would do:
对于您的具体示例，您将执行以下操作：
import pandas as pd
import numpy as np
df = pd.DataFrame(
np.array([[1,2,3,0,5],[1,2,3,4,5],[1,1,1,6,1],[1,0,0,0,0]]),
columns=["a","b","c","d","q"],
index = [0,1,2,3]
)
#I want mean of 0, 2, 3 rows for each a, b, d columns
#. a b d
#0 1 1 2
df.iloc[ [0,2,3], [0,1,3] ].mean(axis=0)
Which outputs:
哪些输出：
a 1.0
b 1.0
d 2.0
dtype: float64
Alternatively, to access via column names, first select on those:
或者，要通过列名访问，首先选择那些：
df[ ['a','b','d'] ].iloc[ [0,1,3] ].mean(axis=0)
To answer the second part of your question (from the comments) you can join multiple DataFrames together using pd.concat. It is faster to accumulate the frames in a list and then pass to pd.concatin one go, e.g.
要回答问题的第二部分（来自评论），您可以使用pd.concat. 将帧累积在列表中然后一次性传递给它会更快pd.concat，例如
dfs = []
for ix in idxs:
dfm = df.iloc[ [0,2,3], ix ].mean(axis=0)
dfs.append(dfm)
dfm_summary = pd.concat(dfs, axis=1) # Stack horizontally
