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()
dict_map = df_1.groupby(‘pos’)[‘value1’].apply(list).to_dict()
本文为原创作品，欢迎分享朋友圈
