Python-科学计算-pandas-21-DF中2列转为字典

zishendianxia

发布于 2021-09-29 15:52:35

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发布于 2021-09-29 15:52:35

文章被收录于专栏：Python工程师Python工程师

系统：Windows 10 编辑器：JetBrains PyCharm Community Edition 2018.2.2 x64 pandas：1.1.5

这个系列讲讲Python的科学计算及可视化
今天讲讲pandas模块
抽取Df中两列构成一个字典

Part 1：场景描述

已知df1，包括6列，"time", "pos", "value1", "value2", "value3", "value4
抽取其中的pos和value1列构成一个字典

由df生成字典

Part 2：代码

import pandas as pd

dict_1 = {"time": ["2019-11-02", "2019-11-03", "2019-11-04", "2019-11-05",
                   "2019-12-02", "2019-12-03", "2019-12-04", "2019-12-05"],
          "pos": ["A", "A", "C", "D", "E", "E", "G", "H"],
          "value1": [10, 20, 30, 40, 50, 60, 70, 80],
          "value2": [100, 200, 300, 400, 500, 600, 700, 800],
          "value3": [50, 20, 30, 90, 50, 60, 80, 80],
          "value4": [10, 30, 90, 40, 60, 60, 70, 80]}

df_1 = pd.DataFrame(dict_1, columns=["time", "pos", "value1", "value2", "value3", "value4"])
print("\n", "df_1", "\n", df_1, "\n")

dict_map = df_1.groupby('pos')['value1'].apply(list).to_dict()

print(dict_map)

print("\n分步骤")
step_1 = df_1.groupby('pos')
print("\n步骤1")
print(step_1)
print(type(step_1))
print(list(step_1))
print(list(step_1)[0])

step_2 = step_1['value1']
print("\n步骤2")
print(step_2)
print(type(step_2))
print(list(step_2))
print(list(step_2)[0])

step_3 = step_2.apply(list)
print("\n步骤4")
print(step_3)
print(type(step_3))

step_4 = step_3.to_dict()
print("\n步骤3")
print(step_4)
print(type(step_4))

代码截图

Part 3：输出结果

Part 4：部分代码解读

dict_map = df_1.groupby('pos')['value1'].apply(list).to_dict()
- dict_map = df_1.groupby(字典键对应列名)[字典值对应列名].apply(字典值组织方式).to_dict()
将字典值组织方式改为集合，dict_map = df_1.groupby('pos')['value1'].apply(set).to_dict()，结果如下，修改了一下数据源，可以实现去重的效果。同样的数据源两种方式差别如下

dict_map = df_1.groupby(‘pos’)[‘value1’].apply(set).to_dict()