经理和员工ID的python层次结构 [英] python hierarchy from manager and employee id

问题描述

我有一个包含两列的csv:employee id 'eid'和经理的employee id 'mid'.试图获取python代码，此代码将为每位员工添加一列，以显示从经理到CEO的经理的员工ID. CEO的雇员ID为1.最终，我想将结果写回csv.

I have a csv with two columns: employee id 'eid' and manager's employee id 'mid'. Trying to get python code that will, for each employee, add columns showing employee id's of the manager all the way to CEO. CEO has a an employee id of 1. Ultimately I want to write the result back to a csv.

所以数据看起来像:

eid,    mid
111,    112
113,    112
112,    114
114,    115
115,    1

我期望输出看起来像这样.请注意，虽然没有一个员工拥有超过4个级别的经理，但是我还想学习动态命名列的python.

I am expecting output that looks like this. Note that while no employee will have more than 4 levels of managers, but i would like to also learn python that names columns dynamically.

eid,    mid,    l2mid   l3mid   l4mid
111,    112,    114,    115,    1
113,    112,    114,    115,    1
112,    114,    115,    1   
114,    115,    1       
115,    1           

我对编码非常陌生，他试图自学，但不断陷入困境.我的问题: 1)我试图使用for语句，该语句在给定行中包含mid，然后发现该经理的经理，依此类推，直到我到达CEO.我一直在尝试以下方法:

I am very new to coding, and trying to teach myself but keep getting stuck. My questions: 1) I was trying to use a for statement that took mid in a given row, then found that that manager's manager, and so on, until i reached the CEO. I have been trying along these lines:

df = pd.read_csv('employee.csv') 
if mid =! 1 
for i in df:
    df.['l2mid'] = df.loc[df.eid == [i], [mid]]

也许我正在倒退，我应该尝试按经理对所有员工进行分组吗?该代码有何不同?

Maybe I'm approaching this backwards and I should try grouping all employees by manager? How would that code be different?

我在 C#和树和

I have seen solutions in C# and sql, and i've seen solutions that build trees and json. I really appreciate any help and encouragement.

更新:下一步是添加国家/地区列-请参阅:

Update: next step was to add a country column - see: entry here

推荐答案

我相信有更好的解决方案，但是可以.我把零填满了.

I believe there is a better solution, but this works. I filled empty with zeros.

a = []
for index, row in df.iterrows():
    res = df[df['eid']==row['mid']]['mid'].values
    a.append(0 if not res else res[0])
df['l2mid'] = a

a = []
for index, row in df.iterrows():
    res = df[df['eid']==row['l2mid']]['mid'].values
    a.append(0 if not res else res[0])
df['l3mid'] = a

a = []
for index, row in df.iterrows():
    res = df[df['eid']==row['l3mid']]['mid'].values
    a.append(0 if not res else res[0])
df['l4mid'] = a

df
# output :
# eid   mid l2mid   l3mid   l4mid
# 0 111 112 114 115 1
# 1 113 112 114 115 1
# 2 112 114 115 1   0
# 3 114 115 1   0   0
# 4 115 1   0   0   0

您可以为例程定义一个函数.

You can define a function for routines.

def search_manager(target_column, new_column):
    a = []
    for index, row in df.iterrows():
        res = df[df['eid']==row[target_column]]['mid'].values
        a.append(0 if not res else res[0])
    df[new_column] = a

search_manager('mid', 'l2mid')
search_manager('l2mid', 'l3mid')
search_manager('l3mid', 'l4mid')

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