彙總計算一個熊貓數據幀

Question

我有一個DF，看起來像底部（摘錄，有4個地區和日期，每個季度擴大）

我想創建一個DF（按地區）只用差異在最新日期與前一個季度和前一年（同一季度）之間

此時，兩個區域和Quradate都是索引。

，所以我想是這樣（沒有真正的接近）：

(['region'] ['Quradate'][-1:-1])-(['region'] ['Quradate'][-2:-2]) 
& (['region'] ['Quradate'][-1:-1])-(['region'] ['Quradate'][-5:-5]) 

，所以我最終會與上一季度每區兩行與分數的差值1日（實際上有5分）第二個與前一年不同。

卡住...

                Score1  Score2 
region           Quradate   
North_Central-Birmingham-Tuscaloosa-Anniston 2010-01-15    47   50 
              2010-04-15    45   60 
              2010-07-15    45   40 
              2010-10-15    42   43 
              2011-01-15    46   44 
              2011-04-15    45   45 
              2011-07-15    45   45 
              2011-10-15    43   46 
              2012-01-15    51   55 
              2012-04-15    53   56 
              2012-07-15    51   57 
              2012-10-15    52   58 
              2013-01-15    50   50 
              2013-04-15    55   55 
              2013-07-15    55   56 
              2013-10-15    51   66 
North_Huntsville-Decatur-Florence   2010-01-15    55   55 

Answer 1

在這裏看到的分辨率和討論：Selecting a new dataframe via a multi-indexed frame in Pandas using index names

基本上所有你需要的是從前期

df.groupby(level='region').apply(lambda x: x.diff().iloc[-1]) 

一個diff和從一年前（4個季度）

df.groupby(level='region').apply(lambda x: x.diff(4).iloc[-1]) 

一個diff

Answer 2

我覺得你有些在正確的軌道上。在我看來，我會做一個函數來計算你正在查找的兩個值並返回一個數據框。像下面這樣：

def find_diffs(region): 
    score_cols = ['Score1', 'Score2'] 

    most_recent_date = region.Quradate.max() 
    last_quarter = most_recent_date - datetime.timedelta(365/4) # shift by 4 months 
    last_year = most_recent_date - datetime.timedelta(365) # shift by a year 

    quarter_score_diff = region[region.Quradate == most_recent_date OR region.Quradate == last_quarter)].diff() 
    quarter_score_diff['id'] = 'quarter_diff' 

    year_score_diff = region[region.Quradate == most_recent_date OR region.Quradate == last_year)].diff() 
    year_score_diff['id'] = 'year_diff' 

    df_temp = quarter_score_diff.append(year_score_diff) 
    return df_temp 

然後，您可以：

DF.groupby(['region']).apply(find_diffs) 

的結果將是一個DF索引通過區域與列各分差和一個額外的列標識每一行作爲一個季度年度差異。

Answer 3

編寫一個函數然後用於groupby絕對是一個選項，另一個很容易做的事情是在組中使用數據列表並使用indeces來進行計算，這是由於常規您的數據的間隔性（並記住，這隻有在數據定期間隔時纔有效）。這種方法必須真正處理日期。首先，我會重新編制，使區域出現在數據幀爲列，那麼我會做到以下幾點：

#First I create some data 
Dates = pd.date_range('2010-1-1', periods = 14, freq = 'Q') 
Regions = ['Western', 'Eastern', 'Southern', 'Norhtern'] 
df = DataFrame({'Regions': [elem for elem in Regions for x in range(14)], \ 
      'Score1' : np.random.rand(56), 'Score2' : np.random.rand(56), 'Score3' : np.random.rand(56), \ 
      'Score4' : np.random.rand(56), 'Score5' : np.random.rand(56)}, index = list(Dates)*4) 

# Create a dictionary to hold your data 
SCORES = ['Score1', 'Score2', 'Score3', 'Score4', 'Score5'] 
ValuesDict = {region : {score : [int(), int()] for score in SCORES} for region in df.Regions.unique()} 

#This dictionary will contain keys that are your regions, and these will correspond to a dictionary that has keys that are your scores and those correspond to a list of which the fisrt element is the most recent - last quarter calculation, and the second is the most recent - last year calcuation. 

#Now group the data 
dfGrouped = df.groupby('Regions') 

#Now iterate through the groups creating lists of the underlying data. The data that is at the last index point of the list is by definition the newest (due to the sorting when grouping) and the obervation one year previous to that is - 4 index points away. 

for group in dfGrouped: 
    Score1List = list(group[1].Score1) 
    Score2List = list(group[1].Score2) 
    Score3List = list(group[1].Score3) 
    Score4List = list(group[1].Score4) 
    Score5List = list(group[1].Score5) 
    MasterList = [Score1List, Score2List, Score3List, Score4List, Score5List] 
    for x in xrange(1, 6): 
     ValuesDict[group[0]]['Score' + str(x)][0] = MasterList[x-1][-1] - MasterList[x-1][-2] 
     ValuesDict[group[0]]['Score' + str(x)][1] = MasterList[x-1][-1] - MasterList[x-1][-5] 

ValuesDict 

它是一種有點令人費解，但是這是我經常接近這些類型的問題的方式。值字典包含您需要的所有數據，但我很難將其納入數據框。