自动彩色海洋堆积条形图

Question

我正在尝试在seaborn中创建shown here的堆叠条形图。

import seaborn as sns
import matplotlib.pyplot as plt
sns.set(style="whitegrid")

# Initialize the matplotlib figure
f, ax = plt.subplots(figsize=(6, 15))

# Load the example car crash dataset
crashes = sns.load_dataset("car_crashes").sort_values("total", ascending=False)

# Plot the total crashes
sns.set_color_codes("pastel")
sns.barplot(x="total", y="abbrev", data=crashes,
            label="Total", color="b")

# Plot the crashes where alcohol was involved
sns.set_color_codes("muted")
sns.barplot(x="alcohol", y="abbrev", data=crashes,
            label="Alcohol-involved", color="b")

# Add a legend and informative axis label
ax.legend(ncol=2, loc="lower right", frameon=True)
ax.set(xlim=(0, 24), ylabel="",
       xlabel="Automobile collisions per billion miles")
sns.despine(left=True, bottom=True)

我在代码中注意到在每个barplot中手动设置了颜色。这看起来很乏味，而且我知道seaborn有一些很棒的色泽可以利用。制作如上所示的堆叠条形图时，如何为每个系列（堆叠）自动设置颜色？

预先感谢

编辑：

为回答以下问题，下面是一个堆叠图形的示例，当我没有手动定义每个系列的颜色时，都使用了我用来实现该颜色的代码。

f, ax = plt.subplots()

xtics = df.index.astype('str')
sns.color_palette("Set1", n_colors=6, desat=.5)

sns.barplot(xtics, df["Series1"], label="Series1")
sns.barplot(xtics, df["Series2"], bottom=df["Series1"], label="Series2")
sns.barplot(xtics, df["Series3"], bottom=df["Series 2"], label="Series3")
sns.barplot(xtics, df["Series4"], bottom=df["Series3"], label="Series4")
sns.barplot(xtics, df["Series5"], bottom=df["Series4"], label = "Series5")
sns.barplot(xtics, df["Series6"], bottom=df["Series5"], label = "Series6")

ax.legend(ncol=1, frameon=True, loc='upper left',  bbox_to_anchor=(1, 0.5))

ax.set(title="Different Color for Each Bars, Same Color for each series", ylabel="YAxis", xlabel="XAxis")
sns.despine(left=True, bottom=True)

以下是我所追求的： 实际的颜色并不重要，我只希望它们对于每个系列都不同，而无需自己手动选择每种颜色。请注意以下代码中的color=。

f, ax = plt.subplots()

xtics = df.index.astype('str')
sns.color_palette("Set1", n_colors=6, desat=.5)

sns.barplot(xtics, df["Series1"], label="Series1", color="#000000")
sns.barplot(xtics, df["Series2"], bottom=df["Series1"], label="Series2", color="#004949")
sns.barplot(xtics, df["Series3"], bottom=df["Series 2"], label="Series3", color="#009292")
sns.barplot(xtics, df["Series4"], bottom=df["Series3"], label="Series4", color="#ff6db6")
sns.barplot(xtics, df["Series5"], bottom=df["Series4"], label = "Series5", color="#490092")
sns.barplot(xtics, df["Series6"], bottom=df["Series5"], label = "Series6", color="#ffb6db")

ax.legend(ncol=1, frameon=True, loc='upper left',  bbox_to_anchor=(1, 0.5))

ax.set(title="Different Color for Each Bars, Same Color for each series", ylabel="YAxis", xlabel="XAxis")
sns.despine(left=True, bottom=True)

Answer 1

我只使用颜色列表并使用循环，或使用所述列表的迭代器，如下所示：

np.random.seed(1234)
N_series = 6
N_bars = 4
xticks = ['2016', '2017', '2018', '2019']
df = pd.DataFrame({f'Series{i+1}': np.random.randint(1,10,size=(N_bars,)) for i in range(N_series)}, index=xticks)

colors = iter(sns.color_palette('Set1', n_colors=N_series, desat=.75))

fig, ax = plt.subplots()
ax.bar(xticks, df["Series1"], bottom=0, label="Series1", color=next(colors))
ax.bar(xticks, df["Series2"], bottom=df["Series1"], label="Series2", color=next(colors))
ax.bar(xticks, df["Series3"], bottom=df["Series2"]+df["Series1"], label="Series3", color=next(colors))
ax.bar(xticks, df["Series4"], bottom=df["Series3"]+df["Series2"]+df["Series1"], label="Series4", color=next(colors))
ax.bar(xticks, df["Series5"], bottom=df["Series4"]+df["Series3"]+df["Series2"]+df["Series1"], label="Series5", color=next(colors))
ax.bar(xticks, df["Series6"], bottom=df["Series5"]+df["Series4"]+df["Series3"]+df["Series2"]+df["Series1"], label="Series6", color=next(colors))

ax.legend(ncol=1, frameon=True, loc='upper left',  bbox_to_anchor=(1, 0.5))
ax.set(title="Same Color for each series", ylabel="YAxis", xlabel="XAxis")
sns.despine(left=True, bottom=True)

PS

• 当您似乎没有以与plt.bar()不同的方式使用seaborn时，为什么要使用seaborn？在我的示例中，我直接使用了bar()，但是如果替换为sns.barplot()
，则会得到相同的输出
• 您在bottom=参数中犯了一个错误，必须提供以下所有小节的和才能获得所需的输出