python - 如何在 python 中迭代给定日期的几个小时？

Question

我有以下温度读数的时间序列数据:

DT                Temperature
01/01/2019 0:00     41
01/01/2019 1:00     42
01/01/2019 2:00     44
......
01/01/2019 23:00    41
01/02/2019 0:00     44

我正在尝试编写一个函数来比较给定日期每小时的温度变化。任何大于 3 的更改都会增加 quickChange 计数器。像这样:

def countChange(day):
    for dt in day:
        if dt+1 - dt > 3: quickChange = quickChange+1

我可以调用该函数一天，例如:countChange(df.loc['2018-01-01'])

Answer 1

使用Series.diff通过 3 进行比较并通过 sum 计算 True 的值:

np.random.seed(2019)

rng = (pd.date_range('2018-01-01', periods=10, freq='H').tolist() +
      pd.date_range('2018-01-02', periods=10, freq='H').tolist())
df = pd.DataFrame({'Temperature': np.random.randint(100, size=20)}, index=rng)  
print (df)
                     Temperature
2018-01-01 00:00:00           72
2018-01-01 01:00:00           31
2018-01-01 02:00:00           37
2018-01-01 03:00:00           88
2018-01-01 04:00:00           62
2018-01-01 05:00:00           24
2018-01-01 06:00:00           29
2018-01-01 07:00:00           15
2018-01-01 08:00:00           12
2018-01-01 09:00:00           16
2018-01-02 00:00:00           48
2018-01-02 01:00:00           71
2018-01-02 02:00:00           83
2018-01-02 03:00:00           12
2018-01-02 04:00:00           80
2018-01-02 05:00:00           50
2018-01-02 06:00:00           95
2018-01-02 07:00:00            5
2018-01-02 08:00:00           24
2018-01-02 09:00:00           28

---

#if necessary create DatetimeIndex if DT is column
df = df.set_index("DT")

def countChange(day):
    return (day['Temperature'].diff() > 3).sum()

print (countChange(df.loc['2018-01-01']))
4

print (countChange(df.loc['2018-01-02']))
9