如何在 x 轴上绘制带有日期时间的线性回归

问题描述

我的 DataFrame 对象看起来像

 数量日期2014-01-06 12014-01-07 12014-01-08 42014-01-09 12014-01-14 1

我想要一种散点图，时间沿 x 轴，数量在 y 上，用一条线穿过数据来引导观察者的视线.如果我使用 pandas plot df.plot(style="o") 这不太正确，因为该行不存在.我想要

用日期轴上的序数绘制图

ax = seaborn.regplot(数据=df，x='date_ordinal',y='金额',)# 收紧坐标轴以保持美观ax.set_xlim(df['date_ordinal'].min() - 1, df['date_ordinal'].max() + 1)ax.set_ylim(0, df['amount'].max() + 1)

用漂亮、可读的日期替换序数 X 轴标签

ax.set_xlabel('date')new_labels = [date.fromordinal(int(item)) for item in ax.get_xticks()]ax.set_xticklabels(new_labels)

哒哒！

My DataFrame object looks like

            amount
date    
2014-01-06  1
2014-01-07  1
2014-01-08  4
2014-01-09  1
2014-01-14  1

I would like a sort of scatter plot with time along the x-axis, and amount on the y, with a line through the data to guide the viewer's eye. If I use the pandas plot df.plot(style="o") it's not quite right, because the line is not there. I would like something like the examples here.

解决方案

note: this has a lot in common with Ian Thompson's answer but the approach is different enough to have it be a separate answer. I use the DataFrame format provided in the question and avoid changing the index.

Seaborn and other libraries don't deal as well with datetime axes as you might like them to. Here's how I'd work around it:

Start by adding a column of date ordinals

Seaborn will deal better with these than with dates. This is a handy trick for doing all kind of mathy things with dates and libraries that don't love dates.

from datetime import date

df['date_ordinal'] = pd.to_datetime(df['date']).apply(lambda date: date.toordinal())

Make a plot with the ordinals on the date axis

ax = seaborn.regplot(
    data=df,
    x='date_ordinal',
    y='amount',
)
# Tighten up the axes for prettiness
ax.set_xlim(df['date_ordinal'].min() - 1, df['date_ordinal'].max() + 1)
ax.set_ylim(0, df['amount'].max() + 1)

Replace the ordinal X-axis labels with nice, readable dates

ax.set_xlabel('date')
new_labels = [date.fromordinal(int(item)) for item in ax.get_xticks()]
ax.set_xticklabels(new_labels)

ta-daa!