Python 数据分析不需要任何东西

编写一个名为proportion_of_education的函数，返回数据集中母亲教育程度低于高中（12）和大学学位的儿童比例

此函数应以以下形式返回字典（使用正确的数字，不要对数字进行四舍五入）：

我复制并尝试使用的代码如下

def proportion_of_education():
    # your code goes here
    # YOUR CODE HERE
    # raise NotImplementedError()
    import pandas as pd
    import numpy as np
    df = pd.read_csv("assests/NISPUF17.csv", index_col=0)
    EDUS=df['EDUC1']
    edus=np.sort(EDUS.values)
    poe={"less than high school":0,
        "high school":0,
        "more than high school but not college":0,
        "college":0}
    n=len(edus)
    poe["less than high school"]=np.sum(edus==1)/n
    poe["high school"]=np.sum(edus==2)/n
    poe["more than high school but not college"]=np.sum(edus==3)/n
    poe["college"]=np.sum(edus==4)/n
    return poe

应该打印出来

{"less than high school":0.2,
    "high school":0.4,
    "more than high school but not college":0.2,
    "college":0.2}

然而，笔记本什么也不扔。这不是一个错误，尽管我使用不同的路径多次运行此操作，但绝对不会打印任何内容。有什么问题。

你应该使用

len（edus==1）

而不是

np.sum

，与其他类似的

df=pd.read\u csv（“assests/NISPUF17.csv”，index\u col=0）

你必须用下面的行替换这一行，因为你写的资产拼写错误

df=pd.read\u csv（“assets/NISPUF17.csv”，index\u col=0）

正确的代码是：

import pandas as pd
import numpy as np

def proportion_of_education():
    
    df = pd.read_csv("assets/NISPUF17.csv", index_col = 0)
    
    EDUS = df['EDUC1']
    edus = np.sort(EDUS.values)
    
    poe = {"less than high school": 0,
        "high school": 0,
        "more than high school but not college": 0,
        "college": 0}
    n = len(edus)
    
    poe["less than high school"] = np.sum(edus == 1)/n
    poe["high school"] = np.sum(edus == 2)/n
    poe["more than high school but not college"] = np.sum(edus == 3)/n
    poe["college"] = np.sum(edus == 4)/n
    
    return poe
    raise NotImplementedError()

作为pd进口熊猫 将numpy作为np导入

定义教育的比例（）

---

教育的比例（）

在问题下方的第一个单元格中键入此项以读取数据

import pandas as pd
df=pd.read_csv('assets/NISPUF17.csv',index_col=0)
df

在下一个单元格中

def proportion_of_education():
    # your code goes here
    cat=pd.value_counts(df['EDUC1'])
    total=sum(cat)
    a_dict=dict(cat)
    s=pd.Series(a_dict)
    f=lambda x: x/total
    s=s.apply(f)
    s_dict=dict(s)
    s_dict['less than high school'] = s_dict.pop(1)
    s_dict['high school'] = s_dict.pop(2)
    s_dict['more than high school but not college'] = s_dict.pop(3)
    s_dict['college'] = s_dict.pop(4)
    return s_dict

    raise NotImplementedError()

proportion_of_education()

要进行检查，请在下一个单元格中键入此项

def proportion_of_education():
    # your code goes here
    cat=pd.value_counts(df['EDUC1'])
    total=sum(cat)
    a_dict=dict(cat)
    s=pd.Series(a_dict)
    f=lambda x: x/total
    s=s.apply(f)
    s_dict=dict(s)
    s_dict['less than high school'] = s_dict.pop(1)
    s_dict['high school'] = s_dict.pop(2)
    s_dict['more than high school but not college'] = s_dict.pop(3)
    s_dict['college'] = s_dict.pop(4)
    return s_dict

    raise NotImplementedError()

proportion_of_education()

这个问题的最后一个单元在下面

assert type(proportion_of_education())==type({}), "You must return a dictionary."
assert len(proportion_of_education()) == 4, "You have not returned a dictionary with four items in it."
assert "less than high school" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "high school" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "more than high school but not college" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "college" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."

在“EDUC1”列中，将有1 2 3 4

=>1-不到高中

=>2-高中

=>3-超过高中，但不超过大学

=>4-大学

lambda函数用于计算所有类别的比例

---

.pop（）方法用于将1重命名为低于高中，类似地，对于其他类别

您可以更好地创建词典。可能类似于

poe=df['EDU'].value_计数（normalize=True）.to_dict（）；返回poe这里看起来也只有断言，但没有“打印”语句（如果你想“打印”一些东西，你需要有这些语句）。在正常的脚本/程序中，你必须使用
print（）
来显示任何东西-
print（教育的比例（））
或者分两步
结果=教育的比例（）
打印（结果）

。在记事本中，您也可以使用

打印（…）

，或者至少您必须在没有任何其他元素的情况下执行函数-

教育的比例（）

，笔记本应该自动显示单元格中最后一个函数的结果。但您使用的是

教育的比例（）

内置

断言

，从函数中获取结果，这样笔记本就不会显示任何内容。

assert type(proportion_of_education())==type({}), "You must return a dictionary."
assert len(proportion_of_education()) == 4, "You have not returned a dictionary with four items in it."
assert "less than high school" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "high school" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "more than high school but not college" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."
assert "college" in proportion_of_education().keys(), "You have not returned a dictionary with the correct keys."