pandas 我可以使用 seaborn 在 x 轴上绘制带有日期时间的线性回归吗？
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原文地址: http://stackoverflow.com/questions/29308729/
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Can I plot a linear regression with datetimes on the x-axis with seaborn?
提问by theQman
My DataFrame object looks like
我的 DataFrame 对象看起来像
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 panadas plot df.plot(style="o")it's not quite right, because the line is not there. I would like something like the examples here.
我想要一种散点图，时间沿 x 轴，数量在 y 上，有一条穿过数据的线来引导观众的眼睛。如果我使用 panadas 情节df.plot(style="o")它不太正确，因为线不在那里。我想要类似这里的例子。
回答by waterproof
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.
注意：这与 Ian Thompson 的答案有很多共同点，但该方法的不同之处足以让它成为一个单独的答案。我使用问题中提供的 DataFrame 格式并避免更改索引。
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:
Seaborn 和其他库并没有像您希望的那样处理日期时间轴。这是我解决它的方法：
Start by adding a column of date ordinals
首先添加一列日期序数
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.
Seaborn 会比处理日期更好地处理这些问题。这是一个方便的技巧，可以用不喜欢日期的日期和库来做各种数学运算。
df['date_ordinal'] = pd.to_datetime(df['date']).apply(lambda date: date.toordinal())
Make a plot with the ordinals on the date axis
用日期轴上的序数绘制一个图
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
用漂亮、可读的日期替换顺序 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)
ta-daa!
哒哒！
回答by Ian Thompson
Since Seaborn has trouble with dates, I'm going to create a work-around. First, I'll make the Date column my index:
由于 Seaborn 在约会方面遇到麻烦，我将创建一个解决方法。首先，我将日期列作为我的索引：
# Make dataframe
df = pd.DataFrame({'amount' : [1,
1,
4,
1,
1]},
index = ['2014-01-06',
'2014-01-07',
'2014-01-08',
'2014-01-09',
'2014-01-14'])
Second, convert the index to pd.DatetimeIndex:
其次，将索引转换为 pd.DatetimeIndex：
# Make index pd.DatetimeIndex
df.index = pd.DatetimeIndex(df.index)
And replace the original with it:
并用它替换原来的：
# Make new index
idx = pd.date_range(df.index.min(), df.index.max())
Third, reindex with the new index (idx):
第三，使用新索引（idx）重新索引：
# Replace original index with idx
df = df.reindex(index = idx)
This will produce a new dataframe with NaN values for the dates you don't have data:
这将为您没有数据的日期生成一个具有 NaN 值的新数据框：
Fourth, since Seaborn doesn't play nice with dates and regression lines I'll create a row count column that we can use as our x-axis:
第四，由于 Seaborn 不能很好地处理日期和回归线，我将创建一个行数列，我们可以将其用作我们的 x 轴：
# Insert row count
df.insert(df.shape[1],
'row_count',
df.index.value_counts().sort_index().cumsum())
Fifth, we should now be able to plot a regression line using 'row_count' as our x variable and 'amount' as our y variable:
第五，我们现在应该能够使用 'row_count' 作为我们的 x 变量和 'amount' 作为我们的 y 变量来绘制回归线：
# Plot regression using Seaborn
fig = sns.regplot(data = df, x = 'row_count', y = 'amount')
Sixth, if you would like the dates to be along the x-axis instead of the row_count you can set the x-tick labels to the index:
第六，如果您希望日期沿着 x 轴而不是 row_count，您可以将 x-tick 标签设置为索引：
# Change x-ticks to dates
labels = [item.get_text() for item in fig.get_xticklabels()]
# Set labels for 1:10 because labels has 11 elements (0 is the left edge, 11 is the right
# edge) but our data only has 9 elements
labels[1:10] = df.index.date
# Set x-tick labels
fig.set_xticklabels(labels)
# Rotate the labels so you can read them
plt.xticks(rotation = 45)
# Change x-axis title
plt.xlabel('date')
plt.show();
Hope this helps!
希望这可以帮助！
