我有像这样的DataFrame
df = pd.DataFrame({
'comments': {0: 0, 1: 1, 2: 47, 3: 102, 4: 230},
'content_len': {0: 4305, 1: 7344, 2: 8431, 3: 5662, 4: 3706},
'day': {0: 1, 1: 1, 2: 1, 3: 2, 4: 2},
'dayofweek': {0: 2, 1: 2, 2: 2, 3: 3, 4: 3},
'domain': {0: 'habrahabr.ru',
1: 'habrahabr.ru',
2: 'habrahabr.ru',
3: 'habrahabr.ru',
4: 'geektimes.ru'},
'favs': {0: 0, 1: 1, 2: 72, 3: 36, 4: 6},
'post_id': {0: 18284, 1: 18285, 2: 18286, 3: 18291, 4: 18294},
'views': {0: 236, 1: 353, 2: 1200, 3: 5700, 4: 1400},
'votes_minus': {0: 0.0, 1: 0.0, 2: 5.0, 3: 3.0, 4: 15.0},
'votes_plus': {0: 0.0, 1: 1.0, 2: 45.0, 3: 72.0, 4: 73.0},
'year_month': {0: datetime.strptime('2008-01-01', '%Y-%m-%d'),
1: datetime.strptime('2008-01-01', '%Y-%m-%d'),
2: datetime.strptime('2008-02-01', '%Y-%m-%d'),
3: datetime.strptime('2008-02-01', '%Y-%m-%d'),
4: datetime.strptime('2008-03-01', '%Y-%m-%d'),}})
现在我想绘制按'year_month'分组的不同图形,每个域一个图形。
例如文章数量 p>
df[df.domain=='habrahabr.ru'].groupby('year_month').count()[['domain']].rename(columns={'domain':'habrahabr.ru'}).join(
df[df.domain=='geektimes.ru'].groupby('year_month').count()[['domain']].rename(columns={'domain':'geektimes.ru'})).plot()
或表示content_len
df[df.domain == 'habrahabr.ru'].groupby('year_month').mean()[['content_len']].rename(columns={'content_len':'habrahabr.ru'}).astype(int).join(
df[df.domain == 'geektimes.ru'].groupby('year_month').mean()[['content_len']].rename(columns={'content_len':'geektimes.ru'}).astype(int)).plot()
是否有比我给出的更优雅的解决方案?
答案 0 :(得分:2)
所有域名的解决方案:
我认为您可以在groupby
函数中添加新列,然后按unstack
重新整形:
What is the difference between size and count in pandas?
a = df.groupby(['year_month', 'domain']).size().unstack(fill_value=0)
print (a)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0 2
2008-02-01 0 2
2008-03-01 1 0
a.plot()
也可以按sum
,mean
...
b = df.groupby(['year_month', 'domain'])['content_len'].mean().unstack(fill_value=0)
print (b)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0.0 5824.5
2008-02-01 0.0 7046.5
2008-03-01 3706.0 0.0
b.plot()
另一个有点慢的解决方案是pivot_table
:
a = df.pivot_table(index='year_month', columns='domain', aggfunc='size', fill_value=0)
print (a)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0 2
2008-02-01 0 2
2008-03-01 1 0
b = df.pivot_table(index='year_month',
columns='domain',
values='content_len',
aggfunc='mean',
fill_value=0)
print (b)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0 5824.5
2008-02-01 0 7046.5
2008-03-01 3706 0.0
过滤域名的解决方案:
如果只需要过滤某些域,请使用boolean indexing
和isin
进行掩码或query
:
df1 = df[df['domain'].isin(['habrahabr.ru','geektimes.ru'])]
a = df1.groupby(['year_month', 'domain']).size().unstack(fill_value=0)
print (a)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0 2
2008-02-01 0 2
2008-03-01 1 0
df1 = df.query('domain == ["habrahabr.ru", "geektimes.ru"]')
a = df1.groupby(['year_month', 'domain']).size().unstack(fill_value=0)
print (a)
domain geektimes.ru habrahabr.ru
year_month
2008-01-01 0 2
2008-02-01 0 2
2008-03-01 1 0