如何计算 pandas 数据框中每分钟出现的次数 [英] How do I count number of occurrences per minute in a pandas data frame

问题描述

我有一个像这样的熊猫数据框:

I have a pandas data frame like this:

timestamp           status  
2019-01-01 09:00:00 FAILED
2019-01-01 09:00:00 FAILED
2019-01-01 09:00:00 UNKNOWN
2019-01-01 09:00:00 PASSED
2019-01-01 09:00:00 PASSED
2019-01-01 09:01:00 PASSED
2019-01-01 09:01:00 FAILED 

如何将每分钟的数据分组并计算每分钟每个状态的数量以获取此数据框:

How can I group the data per minute and count the number of each status per minute to get this data frame:

timestamp           PASSED FAILED UNKNOWN   
2019-01-01 09:00:00 2      2      1
2019-01-01 09:01:00 1      1      0

推荐答案

方法一:

pd.crosstab(df['timestamp'],df['status'])

status               FAILED  PASSED  UNKNOWN
timestamp                                   
2019-01-01-09:00:00       2       2        1
2019-01-01-09:01:00       1       1        0

<小时>

如果你想要像列一样的时间戳:

---

If you want timestamp like a column :

pd.crosstab(df['timestamp'],df['status'],colnames=[None]).reset_index()

             timestamp  FAILED  PASSED  UNKNOWN
0  2019-01-01-09:00:00       2       2        1
1  2019-01-01-09:01:00       1       1        0

方法 2:

df.groupby(['timestamp','status']).size().unstack(fill_value=0)

<小时>

时间比较:

似乎方法2是最快的.

%%timeit
new_df=pd.crosstab(df['timestamp'],df['status'])
21 ms ± 759 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

<小时>

%%timeit
new_df=df.groupby(['timestamp','status']).size().unstack(fill_value=0)
4.65 ms ± 290 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

<小时>

%%timeit
df2 = (
    df
    .groupby(df['timestamp'].map(lambda x: x.replace(second=0)))['status']
    .value_counts()
    .unstack()
    .fillna(0)
    .astype(int)
    .reset_index()
)

8.5 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

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