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補足を追加
answer
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@@ -55,4 +55,74 @@
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にて実現出来るかと思います。
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> ``astype()`` 以降はデータを分単位表記に変えているだけです
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> ``astype()`` 以降はデータを分単位表記に変えているだけです
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---
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**【追記】**
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休憩時間を処理するサンプル
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```Python
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import pandas as pd
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import datetime
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# 休憩時間(とりあえず適当)
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BREAK_START = datetime.time(9, 27)
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BREAK_END = datetime.time(9, 32)
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# datetime.time 型同士の差を求めるUtility関数
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def time_diff(start_time, end_time):
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return datetime.datetime.combine(datetime.date.today(), end_time) - datetime.datetime.combine(datetime.date.today(), start_time)
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# Groupby.apply() にて呼ばれる関数(各行に時間を求める)
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def calc_product_time(data):
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# 後の処理を行いやすくするために DataFrame化しておく
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tmp_df = pd.DataFrame({'start_time': data.shift(1).dt.time,
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'end_time': data.dt.time,
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'total_time': data.diff()},
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index = data.index)
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#print(tmp_df)
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# 各行に対して休憩時間を計算する
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for idx, row in tmp_df.iterrows():
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# 範囲内に休憩開始・休憩終了時間が含まれる場合
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if ((row.start_time <= BREAK_START) &
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(BREAK_START < row.end_time) &
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(row.start_time <= BREAK_END) &
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(BREAK_END < row.end_time)):
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tmp_df.loc[idx, 'break_time'] = time_diff(BREAK_START, BREAK_END)
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# 範囲内に休憩開始時間のみ含まれる場合
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elif ((row.start_time <= BREAK_START) &
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(BREAK_START < row.end_time) &
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(BREAK_END >= row.end_time)):
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tmp_df.loc[idx, 'break_time'] = time_diff(BREAK_START, row.end_time)
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# 範囲内に休憩終了時間のみ含まれる場合
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elif ((row.start_time > BREAK_START) &
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(row.start_time <= BREAK_END) &
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(BREAK_END < row.end_time)):
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tmp_df.loc[idx, 'break_time'] = time_diff(row.start_time, BREAK_END)
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# 休憩時間内に、範囲がすべて含まれる場合
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elif ((row.start_time > BREAK_START) &
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(BREAK_END >= row.end_time)):
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tmp_df.loc[idx, 'break_time'] = time_diff(row.start_time, row.end_time)
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# その他(範囲内に休憩なし)
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else:
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tmp_df.loc[idx, 'break_time'] = datetime.timedelta(0)
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tmp_df['product_time'] = tmp_df['total_time'] - tmp_df['break_time']
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#print(tmp_df)
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return tmp_df['product_time']
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df = pd.read_csv('data.csv', parse_dates={'datetime': ['yyyymmdd', 'hhmm']})
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df['min_per_product'] = df.groupby(['id', df['datetime'].dt.date])['datetime'].apply(calc_product_time)
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print(df)
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```
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