[英]Pandas: Creating a list based on the differences between 2 series
I am writing a custom error message when 2 Pandas series are not equal and want to use '<' to point at the differences.当 2 个 Pandas 系列不相等并且想使用“<”来指出差异时,我正在编写一条自定义错误消息。
Here's the workflow for a failed equality:这是失败的平等的工作流程:
pd.Series([list])
pd.Series([list])
table = pd.concat([list1], [list2]), axis=1
table = pd.concat([list1], [list2]), axis=1
table.columns = ['...', '...']
, table.index = ['...', '...']
table.columns = ['...', '...']
, table.index = ['...', '...']
Current output:当前output:
|Yours|Actual| |你的|实际的|
|1|1| |1|1|
|2|2| |2|2|
|4|3| |4|3|
Desired output:所需的 output:
|Yours|Actual|-| |你的|实际|-|
|1|1|| |1|1||
|2|2|| |2|2||
|4|3|<| |4|3|<|
The naive solution is iterating through each list index and if it's not equal, appending '<' to another list then putting this list into pd.concat()
but I am looking for a method using Pandas. For example,天真的解决方案是遍历每个列表索引,如果它不相等,则将“<”附加到另一个列表,然后将此列表放入
pd.concat()
但我正在寻找一种使用 Pandas 的方法。例如,
error_series = '<' if (abs(yours - actual) >= 1).all(axis=None) else ''
Ideally it would append '<' to a list if the difference between the results is greater than the Margin of Error of 1, otherwise append nothing理想情况下,如果结果之间的差异大于 1 的误差幅度,则 append '<' 到列表,否则 append 没有
Note: Removed tables due to StackOverflow being picky and not letting my post my question注意:由于 StackOverflow 很挑剔并且不让我发布我的问题而删除了表格
You can create the DF and give index and column names in one line:您可以创建 DF 并在一行中给出索引和列名:
import pandas as pd
list1 = [1,2,4]
list2 = [1,2,10]
df = pd.DataFrame(zip(list1, list2), columns=['Yours', 'Actual'])
Create a boolean mask to find the rows that have a too large difference:创建一个 boolean 掩码以查找差异过大的行:
margin_of_error = 1
mask = df.diff(axis=1)['Actual'].abs()>margin_of_error
Add a column to the DF and set the values of the mask as you want:向 DF 添加一列并根据需要设置掩码的值:
df['too_different'] = df.diff(axis=1)['Actual'].abs()>margin_of_error
df['too_different'].replace(True, '<', inplace=True)
df['too_different'].replace(False, '', inplace=True)
output: output:
Yours Actual too_different
0 1 1
1 2 2
2 4 10 <
or you can do something like this:或者你可以这样做:
df = df.assign(diffr=df.apply(lambda x: '<'
if (abs(x['yours'] - x['actual']) >= 1)
else '', axis=1))
print(df)
'''
yours actual diffr
0 1 1
1 2 2
2 4 3 <
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