使用其中一个数据帧作为键,将Python中的数据帧组合到字典中

问题描述 投票:1回答:1

我有3个数据帧,包含每日数据:唯一代码,名称,分数。第1行中的第一个值称为Rank,然后我有日期,Rank下的第一列包含排名编号(第一列用作索引)。

**df1** UNIQUE CODES

Rank 12/8/2017 12/9/2017 .... 1/3/2018
1     Code_1      Code_3       Code_4
2     Code_2      Code_1       Code_2
...
1000  Code_5      Code_6       Code_7

**df2** NAMES

Rank 12/8/2017 12/9/2017 .... 1/3/2018
1     Jon        Maria         Peter
2     Brian      Jon           Maria
...
1000  Chris      Tim           Charles

**df3** SCORES

Rank 12/8/2017 12/9/2017 .... 1/3/2018
1     10           20           30
2     15           10           40
...
1000  25           15           20

期望的输出:

我想将这些数据帧组合成一个字典,使用df1代码作为键,所以它看起来像这样:

dictionary = {'Code_1':[Jon, 20] , 'Code_2':[Brian, 15]}

由于有重复的竞争对手,我需要在所有数据系列中总结他们的分数。因此,在上面的示例中,Jon的Score_1将包含12/8/2017和12/9/2017的分数。

有1000行和26列+索引,所以需要一种方法来捕获它们。我认为嵌套循环可以在这里工作,但没有足够的经验来构建一个有效的循环。

最后,我想按最高分对字典进行排序。请建议使用这种或更直接的方法来解决这些数据并获得分数排名的任何解决方案。

我附上了数据帧的图片,包含名称,代码和分数。

names

codes

scores

我在下面的3个数据帧中使用了提出的解决方案。请注意,hashtags代表代码,名字的玩家和分数的奖杯:

# reshape to get dates into rows
hashtags_reshaped = pd.melt(hashtags, id_vars = ['Rank'], 
                   value_vars = hashtags.columns, 
                   var_name = 'Date', 
                   value_name = 'Code').drop('Rank', axis = 1)

# reshape to get dates into rows
players_reshaped = pd.melt(players, id_vars = ['Rank'], 
                   value_vars = hashtags.columns, 
                   var_name = 'Date', 
                   value_name = 'Name').drop('Rank', axis = 1)

# reshape to get the dates into rows
trophies_reshaped = pd.melt(trophies, id_vars = ['Rank'], 
                   value_vars = hashtags.columns, 
                   var_name = 'Date', 
                   value_name = 'Score').drop('Rank', axis = 1)

# merge the three together. 
# This _assumes_ that the dfs are all in the same order and that all the data matches up.
merged_df = pd.DataFrame([hashtags_reshaped['Date'], 
hashtags_reshaped['Code'], players_reshaped['Name'], 
trophies_reshaped['Score']]).T
print(merged_df)

# group by code, name, and date; sum the scores together if multiple exist for a given code-name-date grouping
grouped_df = merged_df.groupby(['Code', 'Name', 'Date']).sum().sort_values('Score', ascending = False)
print(grouped_df)

summed_df = merged_df.drop('Date', axis = 1) \
.groupby(['Code', 'Name']).sum() \
.sort_values('Score', ascending = False).reset_index()
summed_df['li'] = list(zip(summed_df.Name, summed_df.Score))
print(summed_df)

但我得到一个奇怪的输出:总和分数应该是几百或几千(平均分数是200-300,平均参与频率是4-6倍)。我得到的分数结果是偏离,但他们的匹配代码和名称正确。

summed_df:

0       (MandiBralaX, 996871590076253)  
1              (Arso_C, 9955130513430)  
2               (ThatRainbowGuy, 9946)  
3                         (fabi, 9940)  
4                      (Dogão, 991917)  
5                      (Hierbo, 99168)  
6               (Clyde, 9916156180128)  
7      (.A.R.M.I.N., 9916014310187143)  
8                (keftedokofths, 9900)  
9                   (⚽AngelSosa⚽, 990)  
10                       (Totoo98, 99)  

group_df:

          Code             Name             Score  \
0       #JL2J02LY      MandiBralaX   996871590076253   
1       #80JQ90VC           Arso_C     9955130513430   
2       #9GGC2CUQ   ThatRainbowGuy              9946   
3       #8LL989QV             fabi              9940   
4        #9PPC89L            Dogão            991917   
5      #2JPLQ8JP8           Hierbo             99168
python loops dictionary dataframe ranking
1个回答
1
投票

这应该可以帮到你。我没有像你指定的那样在最后创建一个字典;虽然您可能需要这种格式,但您最终会得到嵌套的词典或列表,因为每个代码都有1个名称,但可能有很多与之关联的日期和分数。你怎么想要那些记录 - 列表,字典等?

下面的代码返回一个分组的数据帧;您可以将其直接输出到dict(显示),但您可能希望详细指定格式,尤其是在需要有序字典时。 (字典本身没有被排序;如果你真的需要一个有序的字典,你必须from collections import OrderedDict并查看该文档。

import pandas as pd

#create the dfs; note that 'Code' is set up as a string
df1 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['1', '2'], '12/9/2017': ['3', '1']})
df1.set_index('Rank', inplace = True)

# reshape to get dates into rows
df1_reshaped = pd.melt(df1, id_vars = ['Rank'], 
                       value_vars = df1.columns, 
                       var_name = 'Date', 
                       value_name = 'Code').drop('Rank', axis = 1)
#print(df1_reshaped)

# create the second df
df2 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['Name_1', 'Name_2'], '12/9/2017': ['Name_3', 'Name_1']})
df2.set_index('Rank', inplace = True)

# reshape to get dates into rows
df2_reshaped = pd.melt(df2, id_vars = ['Rank'], 
                       value_vars = df1.columns, 
                       var_name = 'Date', 
                       value_name = 'Name').drop('Rank', axis = 1)
#print(df2_reshaped)

# create the third df
df3 = pd.DataFrame({'Rank': [1, 2], '12/8/2017': ['10', '20'], '12/9/2017': ['30', '10']})
df3.set_index('Rank', inplace = True)

# reshape to get the dates into rows
df3_reshaped = pd.melt(df3, id_vars = ['Rank'], 
                       value_vars = df1.columns, 
                       var_name = 'Date', 
                       value_name = 'Score').drop('Rank', axis = 1)
#print(df3_reshaped)

# merge the three together. 
# This _assumes_ that the dfs are all in the same order and that all the data matches up.
merged_df = pd.DataFrame([df1_reshaped['Date'], df1_reshaped['Code'], df2_reshaped['Name'], df3_reshaped['Score']]).T
print(merged_df)

# group by code, name, and date; sum the scores together if multiple exist for a given code-name-date grouping
grouped_df = merged_df.groupby(['Code', 'Name', 'Date']).sum().sort_values('Score', ascending = False)
print(grouped_df)

summed_df = merged_df.drop('Date', axis = 1) \
    .groupby(['Code', 'Name']).sum() \
    .sort_values('Score', ascending = False).reset_index()
summed_df['li'] = list(zip(summed_df.Name, summed_df.Score))
print(summed_df)

未排序的词典:

d = dict(zip(summed_df.Code, summed_df.li))
print(d)

当然,您可以直接制作OrderedDict,并且应该:

from collections import OrderedDict
d2 = OrderedDict(zip(summed_df.Code, summed_df.li))
print(d2)

summed_df

  Code    Name  Score            li
0    3  Name_3     30  (Name_3, 30)
1    1  Name_1     20  (Name_1, 20)
2    2  Name_2     20  (Name_2, 20)

d

{'3': ('Name_3', 30), '1': ('Name_1', 20), '2': ('Name_2', 20)}

d2,已分类:

OrderedDict([('3', ('Name_3', 30)), ('1', ('Name_1', 20)), ('2', ('Name_2', 20))])

这会将您的(姓名,分数)作为元组返回,而不是列表,但是......它应该会有更多的方式。

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