熊猫直方图箱对齐

Question

我有一个看起来像这样的数据框：

train_data_10users = pd.DataFrame({'target':['A','A','B', 'B', 'C'], 'day_of_week':[4,2,4,4,1]})

 target  day_of_week
0   A            4
1   A            2
2   B            4
3   B            4
4   C            1

我想按 day_of_week 为每个目标创建一个计数直方图，即

"A" should have:
0,1,3,5,6:0
2,4:1
"B" should have
0,1,2,3,5,6:0
4:2
"C" should have 1:1, the rest:0

这是显示我想要在直方图上显示的真实数据的数据透视表（注意：fillna）：

pivot = pd.pivot_table(train_data_10users,
                       index=["target"], columns=["day_of_week"], aggfunc='size', fill_value=0)

day_of_week 0   1   2   3   4   5   6
target                          
Ashley  390 328 1078    293 115 0   0
Avril   148 402 273 318 87  104 311
Bill    308 239 105 24  54  7   65
Bob 51  285 72  284 330 0   0

即使 groupby 中可能缺少某些日子，添加适当的 xticks 也能解决问题：

from matplotlib import pyplot as plt
import pandas as pd

fig, axes = plt.subplots(nrows=3, ncols=4, figsize=(16, 10))
for idx, (user, sub_df) in enumerate(
        pd.groupby(train_data_10users[["target", "day_of_week"]], 'target')): 
    ax = axes[idx // 4, idx % 4]
    sub_df.hist(ax=ax, label=user, color=color_dic.get(user), bins=7)
    ax.set_xticks(range(7))
    ax.legend()

但是这些值并不完全对齐/居中，而且位置有点浮动，我认为这取决于每个目标存在/缺失的天数：

更新。 这是根据接受的答案的外观：

fig, axes = plt.subplots(nrows=3, ncols=4, figsize=(16, 10), sharey=True)
...
sub_df.hist(ax=ax, label=user, color=color_dic.get(user), bins=range(8))
ax.set_xticks(range(8))
ax.set_xticks(np.arange(8)+0.5)
ax.set_xticklabels(range(7))

Answer 1

尝试：

fig, axes = plt.subplots(nrows=3, ncols=4, figsize=(16, 10))
for idx, (user, sub_df) in enumerate(
    pd.groupby(train_data_10users[["target", "day_of_week"]], 'target')): 
    ax = axes[idx // 4, idx % 4]

    # note bin is forced to range(7)
    sub_df.hist(ax=ax, label=user, bins=range(7))

    # offset the xticks
    ax.set_xticks(np.arange(7) + .5)

    # name the label accordingly
    ax.set_xticklabels(range(7))

带bins=range(7)输出：