使用 Pandas 进行数据分析没有任何结果答案

【问题标题】：Data Analysis with Pandas throws nothing使用 Pandas 进行数据分析没有任何结果
【发布时间】：2020-12-24 22:52:32
【问题描述】：

编写一个名为 ratio_of_education 的函数，它返回数据集中的孩子的比例，其母亲的教育水平等于低于高中 (12) 和大学学位。

这个函数应该以如下形式返回一个字典（使用正确的数字，不要四舍五入）：

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

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

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."

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}

但是，Notebook 什么也没抛出。不是错误，尽管我使用不同的路径多次运行它，但绝对没有打印出来。可能是什么问题。

【问题讨论】：

您可以更好地创建字典。也许像poe = df['EDU'].value_counts(normalize=True).to_dict(); return poe;
此外，这里似乎只有断言，但没有“打印”语句（如果你想“打印”某些东西，你需要拥有的东西）。
在普通脚本/程序中，您必须使用 print() 来显示任何内容 - print(proportion_of_education()) 或分两步 results = proportion_of_education() print(results)。在笔记本中你也可以使用print(...) 或者至少你必须在没有任何其他元素的情况下执行函数 - proportion_of_education() - 并且笔记本应该自动显示单元格中最后一个函数的结果。但是您在 assert 中使用 proportion_of_education() 会从函数中获取结果，因此 Notebook 没有可显示的内容。

标签： python pandas data-analysis

【解决方案1】：

您应该使用len(edus==1) 而不是np.sum，并且与其他人类似

【讨论】：

【解决方案2】：

df = pd.read_csv("assests/NISPUF17.csv", index_col=0) 您必须将此行替换为下面的行，因为您编写了错误的资产拼写。

df = pd.read_csv("assets/NISPUF17.csv", index_col=0)

【讨论】：

【解决方案3】：

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()

【讨论】：

【解决方案4】：

将熊猫导入为 pd 将 numpy 导入为 np

def ratio_of_education()：

# your code goes here
# YOUR CODE HERE

df = pd.read_csv(r'assets/NISPUF17.csv',index_col=0)
EDUC1 = df['EDUC1']
educ1 = np.sort(EDUC1.values)
poe = {"less than high school": 0,
    "high school": 0,
    "more than high school but not college": 0,
    "college": 0}
total = len(educ1)
poe['less than high school'] = len(df[df['EDUC1'] == 1])/total
poe['high school'] = len(df[df['EDUC1'] == 2])/total
poe['more than high school but not college'] = len(df[df['EDUC1'] == 3])/total
poe['college'] = len(df[df['EDUC1'] == 4])/total
return poe
raise NotImplementedError()

proportion_of_education()

【讨论】：

【解决方案5】：

在问题下方的第一个单元格中键入以读取数据

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()

要检查，请在下一个单元格中输入此内容

proportion_of_education()

这个问题的最后一个单元格在下面

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，会有 1 2 3 4

=> 1 - 低于高中

=> 2 - 高中

=> 3 - 超过高中但不是大学

=> 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 高中以下，其他类别也类似

【讨论】：

【解决方案6】：

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


    proportion_of_education()

【讨论】：

欢迎来到 StackOverflow！感谢您的贡献，虽然此代码可能会回答问题，但包含解释会更有帮助。此外，您有一些缩进错误，因此代码可能无法完全按照发布的那样工作。

【解决方案7】：

简单的解决方案

def proportion_of_education():
    import pandas as pd


    df = pd.read_csv("NISPUF17.csv")

    df = df["EDUC1"]
    lessThanHighSchool = (df[df == 1].count()) / df.count()
    highScool = (df[df == 2].count()) / df.count()
    moreThanHighSchoolNotCollage = (df[df == 3].count()) / df.count()
    collage = (df[df == 4].count()) / df.count()

    dictionary = {"less than high school ":lessThanHighSchool ,
        "high school":highScool,
        "more than high school but not college":moreThanHighSchoolNotCollage,
        "college":collage
        }

    return dictionary

【讨论】：

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