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我用这个问题制作了一个像这样的动画散点图：

import matplotlib.pyplot as plt
import pandas as pd
from matplotlib.collections import PathCollection


df = pd.DataFrame({"x": [1, 2, 6, 4, 5, 6], "y": [1, 4, 36, 16, 25, 36]})
plt.ion()
fig: plt.Figure = plt.figure
ax = fig.subplots()
path_collection: PathCollection = ax.scatter(df.loc[0:2, "x"], df.loc[0:2, "y"])
# Note: I don't use pandas built in DataFrame.plot.scatter function so I can get the PathCollection object to later change the scatterpoints.
fig.canvas.draw()
path_collection.set_offsets([[row.x, row.y] for index, row in df.loc[3:].iterrows()])
# Due to the format of offset (array-like (N,2)) this seems to be the best way to provide the data.

fig.canvas.draw()

这很好用，但我想在 x 轴上有时间，所以我尝试将上面的代码更改为：

import matplotlib.pyplot as plt
import pandas as pd
from matplotlib.collections import PathCollection

df = pd.DataFrame({'time': [pd.Timestamp('2021-02-04 00:00:01'),
                            pd.Timestamp('2021-02-04 00:00:02'),
                            pd.Timestamp('2021-02-04 00:00:10'),
                            pd.Timestamp('2021-02-04 00:00:05'),
                            pd.Timestamp('2021-02-04 00:00:06'),
                            pd.Timestamp('2021-02-04 00:00:08')],
                   'y': [5, 6, 10, 8, 9, 10]})
fig: plt.Figure = plt.figure()
ax = fig.subplots()
sc: PathCollection = ax.scatter(df.loc[0:2, "time"], df.loc[0:2, "y"])
fig.canvas.draw()    
sc.set_offsets([[row.time, row.y] for index, row in df.loc[3:].iterrows()])

fig.canvas.draw()

倒数第二行抛出此错误： TypeError: float() argument must be a string or a number, not 'Timestamp'. 这似乎是由于将其PathCollection存储_offsets为不能包含Timestamp. 所以我想知道，是否有一种解决方法可以用时间轴为散点设置动画？

提前致谢。

4

1 回答 1

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对于任何有同样问题的人，我找到了一个可能远非理想的解决方案，但它确实有效。事实证明，PathCollection 将时间存储为np.float64表示自 0001 年 1 月 1 日以来的天数。而且由于set_offset似乎只能使用大小为 (N, 2) 的类似数组的对象，因此我将自 0001 年 1 月 1pd.Timestamp日以来的天数重新格式化如下：

time = pd.Timestamp('2021-02-04 00:00:01')
sec_since_1970 = time.timestamp()           # pd.Timestamp.timestamp() gives seconds since epox.
days_since_1970 = sec_since_1970/60/60/24   # Seconds to days.
days_since_0001 = days_since_1970 + 719163  # 719163 = number of days from 1/1/0001 until 1/1/1970

在问题中实现这一点给出了以下内容：

df = pd.DataFrame({'time': [pd.Timestamp('2021-02-04 00:00:01'),
                            pd.Timestamp('2021-02-04 00:00:02'),
                            pd.Timestamp('2021-02-04 00:00:10'),
                            pd.Timestamp('2021-02-04 00:00:05'),
                            pd.Timestamp('2021-02-04 00:00:06'),
                            pd.Timestamp('2021-02-04 00:00:08')],
                   'y': [5, 6, 10, 8, 9, 10]})
data = [[719163 + row.time.timestamp()/60/60/24, row.y] for index, row in df.loc[3:].iterrows()]
sc.set_offsets(data)
# Set axis limits
ax.set_ylim(bottom=df.y.loc[3:].min() -1, top=df.y.loc[3:].max()+1)
ax.set_xlim(left= (df.time.loc[3:].min().timestamp()-1)/60/60/24+719163,
            right=(df.time.loc[3:].max().timestamp()+1)/60/60/24+719163)
fig.canvas.draw()

所以这解决了这个问题，它可能不是最有效的方法，但对于小型数据集，它工作得很好。

于 2021-02-08T20:01:21.313 回答