Question

我想使用pandas的下拉菜单来绘制ipywidgets数据帧的子集，但出现一些奇怪的错误。

import ipywidgets as wg
from IPython.display import display
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Make Data Frame
#
df = pd.DataFrame({
    
    "x": np.arange(32),
    "y": np.arange(32),
    "A": [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
    "B": [0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1],
    "C": [0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1]
})

# Make dropdown menus
#
w1 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='A:',
)
#
w2 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='B:',
)
#
w3 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='C:',
)
#
w4 = wg.Dropdown(
    options=['df'],
    value='df',
    description='DF:',
)

# Define plotting function
#
def myPlot(df, a, b, c):
    print(df)
    print(a)
    print(b)
    print(c)
    x = df["x"][df["A"]==a & df["B"]==b & df["C"]==c]
    y = df["y"][df["A"]==a & df["B"]==b & df["C"]==c]   
    plt.scatter(x,y)
    plt.show()

# Plot with interactive dropdown menus
#
wg.interact(myPlot, df=w4, a=w1, b=w2, c=w3)

当我尝试在绘图函数x中定义TypeError: string indices must be integers时发生错误。我认为这与将数据帧放入绘图函数有关，因为绘图函数中的print命令为我提供了A，B和{{1 }}，但显示字符串C。

如何获取我想要的情节？

Answer 1

我不会尝试使用字符串从名称空间中提取变量。

我将包装您的交互器并直接传递数据框：

import ipywidgets as wg
from IPython.display import display
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt


# Make Data Frame
#
df = pd.DataFrame({
    "x": np.arange(32),
    "y": np.arange(32),
    "A": [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
    "B": [0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1],
    "C": [0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1]
})

# Make dropdown menus
#
w1 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='A:',
)
#
w2 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='B:',
)
#
w3 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='C:',
)


# Define plotting function
#
def myPlot(df, a, b, c):
    subset = df.loc[df["A"].eq(a) & df["B"].eq(b) & df["C"].eq(c), :]
    fig, ax = plt.subplots()
    ax.scatter("x", "y", data=subset)
    return fig


def interactive_plotter(df):
    df_widget = wg.fixed(df)
    return wg.interact(myPlot, df=df_widget, a=w1, b=w2, c=w3)

fig = interactive_plotter(df)

要考虑的另一件事是您的链接逻辑语句：

逻辑运算符的操作顺序不是很直观

此声明：

x = df["x"][df["A"]==a & df["B"]==b & df["C"]==c]

评估为：

x = df["x"][df["A"] == (a & df["B"]) == (b & df["C"]==c)]

（或接近它的地方）。

您想要的至少是：

x = df["x"][(df["A"] == a) & (df["B"] == b) & (df["C"] == c)]

但是我认为最好使用.loc访问器：

x = df.loc[(df["A"] == a) & (df["B"] == b) & (df["C"] == c), "x"]

如果您不喜欢所有这些括号，也可以使用.eq()

x = df.loc[df["A"].eq(a) & df["B"].eq(b) & df["C"].eq(c), "x"]

Answer 2

如果您确实希望在不同的数据框之间切换以进行绘制，请将其放入字典中，为每个数据框赋予唯一的字符串键，然后将该字典作为选项传递给下拉列表。


import ipywidgets as wg
from IPython.display import display
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Make Data Frame
#
df = pd.DataFrame({
    
    "x": np.arange(32),
    "y": np.arange(32),
    "A": [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
    "B": [0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1],
    "C": [0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1]
})

choices = {
    'df': df
}

# Make dropdown menus
#
w1 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='A:',
)
#
w2 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='B:',
)
#
w3 = wg.Dropdown(
    options=[0,1],
    value=0,
    description='C:',
)
#
w4 = wg.Dropdown(
    options=['df'],
    value='df',
    description='DF:',
)

# Define plotting function
#
def myPlot(df_name, a, b, c):
    df = choices[df_name]
    x = df.loc[(df["A"]==a) & (df["B"]==b) & (df["C"]==c)]["x"]
    y = df.loc[(df["A"]==a) & (df["B"]==b) & (df["C"]==c)]["y"]  
    plt.scatter(x,y)
    plt.show()

# Plot with interactive dropdown menus
#
wg.interact(myPlot, df_name=w4, a=w1, b=w2, c=w3)

PS。同样，对DataFrame的索引调用的顺序错误。 1）过滤行，然后2）选择所需的列。

df.loc[(df["A"]==a) & (df["B"]==b) & (df["C"]==c)]["x"] # correct order

df['x'].loc[(df["A"]==a) & (df["B"]==b) & (df["C"]==c)] # incorrect order

使用ipywidget绘制子集的熊猫数据框

2 个答案: