使用python和matplotlib的图形直方图 [英] graph histogram using python and matplotlib
问题描述
我有一个2014年至2018年的样本数据,需要绘制直方图才能找到异常值.但是首先,我需要弄清楚2014、2015 ... 2018年有多少个唯一的162个ID,然后再进行规划.我首先存储2014年的data_2014 = data ['DATE'].dt.year == 2014,但是如何找到2014年出现的162个唯一ID中的哪个?非常感谢!
I have a sample data from 2014 through 2018 and need to plot a histogram to find outliers. But first, I need to figure out how many of the unique 162 IDs are in 2014, 2015...2018 and then plot it out. I first store data_2014 = data['DATE'].dt.year == 2014 for year 2014, but how do I find which of the 162 unique IDs occurred in 2014? Thank you so much!
| ID | DATE | VIOLATIONS |
| 0 CHI065 | 2014-07-08 | 65 |
| 1 CHI010 | 2014-07-16 | 56 |
| 2 CHI069 | 2014-07-08 | 10 |
| 3 CHI010 | 2014-07-26 | 101 |
| 4 CHI010 | 2014-07-27 | 92 |
| 5 CHI068 | 2014-08-03 | 20 |
| 17049 CHI040 | 2018-12-22 | 15 |
| 170496 CHI168 | 2018-12-23 | 16 |
| 170497 CHI103 | 2018-12-23 | 8 |
推荐答案
import pandas as pd
df = pd.DataFrame({'date': {0: '26-1-2014', 1: '26-1-2014', 2:'26-1-2015', 3:'30-1-2014'},
'ID': {0:"id12", 1: "id13", 2: "id14", 3: "id12"}, 'violations': {0: 34, 1:3, 2: 45, 3: 15} } )
df['year'] = pd.to_datetime(df.date).dt.strftime('%Y')
每年以字典或数据框形式返回唯一 ID,以便于查找
Return unique Ids per year as dictionary or dataframe for easy lookup
d = df.groupby('year')['ID'].apply(set).to_dict() # as dictionary
d['2014'] #returns unique ids for 2014
以下行创建一个每年具有唯一ID的df.如果您只想知道哪些ID是2014年的一部分,那就很好了.
The following line creates a df with unique IDs per year. This is good if you just want to know which ids are part of 2014.
df_ids = df.groupby('year')['ID'].apply(set).to_frame(name="id_per_year") #as dataframe
您现在可以对年份进行子集化,例如仅获取 2014 年的行
You can now subset on year for example to get only the rows from 2014
df = df.loc[df['year'] == '2014'] # subset for 2014
如果您只想计算2014年的唯一ID,则可以按年份分组并使用nunique()
If you only want to count the unique IDs for 2014 you can groupby year and use nunique()
df_unique = df.groupby('year')['ID'].nunique().to_frame(name="unique_counts")
以下行创建一个包含每年 ID 计数的框架
The following line creates a frame with counts of IDs per year
df_counts = df.groupby('year')['ID'].count().to_frame(name="count")
希望这会有所帮助
这将生成一个表格,其中包含每个 ID 的数量 + 今年的违规总数.
This will generate a table with the number count for each ID + its total number of violations for this year.
import pandas as pd
df = pd.DataFrame({'date': {0: '26-1-2014', 1: '26-1-2014', 2:'26-1-2015', 3:'30-1-2014'},
'ID': {0:"id12", 1: "id13", 2: "id14", 3: "id12"}, 'violations': {0: 34, 1:3, 2: 45, 3: 15} } )
df['year'] = pd.to_datetime(df.date).dt.strftime('%Y')
aggregations = {'ID': 'count', 'violations': 'sum'}
df_agg = df.groupby(['year', 'ID']).agg(aggregations)
corr = df_agg.groupby('year')[['ID', 'violations']].corr() #optional
如果您喜欢每年唯一 ID 的数量,您可以调整聚合和分组
If you like the number of unique IDs per year you can adjust the aggregations and the grouping
aggregations = {'ID': pd.Series.nunique, 'violations': 'sum'}
df_agg = df.groupby('year').agg(aggregations)
您可以像这样绘制散点图.确保在调色板中为每年添加一种颜色.
You can make a scatter plot like this. Make sure to add a color for each year in palette.
import seaborn as sns
sns.scatterplot(df_agg["ID"], df_agg["violations"],hue=df_agg.index.get_level_values("year"),palette=["r", "b"], legend='full')
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