回答編集履歴
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別解を追記
answer
CHANGED
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@@ -21,4 +21,37 @@
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21
21
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4 2018-01-26 b 3 days
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22
22
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5 2018-01-30 b 4 days
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23
23
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"""
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24
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```
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25
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26
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#### 別解：先頭から舐める版
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27
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ユニークidが多く、各id毎の行数が少ない場合は、以下のように先頭から舐めて計算する方が速いかもしれません。
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28
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```Python
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29
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import pandas as pd
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30
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31
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df = pd.DataFrame({'id':['a','a','a','b','b','b'],'date':['2018/01/23','2018/01/24','2018/01/26','2018/01/23','2018/01/26','2018/01/30']},
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32
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columns = ['id','date'])
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33
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df.loc[:,'date'] = pd.to_datetime(df['date'])
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34
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df['delta'] = 0
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35
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36
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prev_id,prev_date = df.loc[0,'id'], df.loc[0,'date']
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37
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for idx,row in df.iterrows():
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38
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cur_id = row['id']
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39
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cur_date = row['date']
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40
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if prev_id != cur_id:
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pass
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else:
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43
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df.loc[idx,'delta'] = cur_date - prev_date
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prev_id = cur_id
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45
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prev_date = cur_date
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46
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print(df)
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48
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"""
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id date delta
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50
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0 a 2018-01-23 0 days 00:00:00
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1 a 2018-01-24 1 days 00:00:00
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2 a 2018-01-26 2 days 00:00:00
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53
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3 b 2018-01-23 0
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54
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4 b 2018-01-26 3 days 00:00:00
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55
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5 b 2018-01-30 4 days 00:00:00
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"""
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24
57
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```
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