具有 95 间隔的 Python 差异回归系数图

Question

我想用 95ci 绘制差异系数图。

这是我的数据框。

example:
x          coef        stderr     ci_lower    ci_upper
t-3        .0005219    .0004025    -.000759    .0018028
t-2        .0008449    .0004205    -.0004935   .0021833
t-1         0             0             0         0
t          .0009092    .0003047    -.0000604   .0018789
t+1        .0010828    .0002889    .0001634    .0020023
t+2        .0013157    .0001613    .0008022    .0018291
t+3        .001452     .0001755    .0008933    .0020106

这是我使用的代码。但它没有用，而且信息量不大

plt.errorbar(x=x, y=coef, yerr=ci_lower(?), color="black", capsize=3,
             linestyle="None",
             marker="s", markersize=7, mfc="black", mec="black")

有没有办法得到这样的回归系数图？

提前致谢

Answer 1

您应该提供数据以及您喜欢的绘图示例。您的数据将是：

df = {'x': {0: 't-3', 1: 't-2', 2: 't-1', 3: 't', 4: 't+1', 5: 't+2', 6: 't+3'},
 'coef': {0: 0.0005219,
  1: 0.0008449,
  2: 0.0,
  3: 0.0009092,
  4: 0.0010828,
  5: 0.0013157,
  6: 0.001452},
 'stderr': {0: 0.0004025,
  1: 0.0004205,
  2: 0.0,
  3: 0.0003047,
  4: 0.0002889,
  5: 0.0001613,
  6: 0.0001755},
 'ci_lower': {0: -0.000759,
  1: -0.0004935,
  2: 0.0,
  3: -6.04e-05,
  4: 0.0001634,
  5: 0.0008022,
  6: 0.0008933},
 'ci_upper': {0: 0.0018028,
  1: 0.0021833,
  2: 0.0,
  3: 0.0018789,
  4: 0.0020023,
  5: 0.0018291,
  6: 0.0020106}}

df = pd.DataFrame(df)

我猜是这样的：

fig, ax = plt.subplots()
df.plot.scatter(x = "coef", y = "x",ax=ax)
for ix in df.index:
    ax.plot(df[['ci_lower','ci_upper']].loc[ix].values, df.loc[ix,['x','x']].values,c="b")