pandas 在python数据框中将月份转换为季度
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Converting month to quarter in python dataframe
提问by Pranav Kansara
I have a column in my data frame denoting month (in the form yyyy-mm). I want to convert that to quarter using pd.Period.
I tried using apply function in below form but it's running too slow. Is there a better way to do this?
I am using :
我的数据框中有一列表示月份（形式为yyyy-mm）。我想使用pd.Period. 我尝试在下面的表单中使用 apply 函数，但它运行速度太慢。有一个更好的方法吗？我在用 ：
hp2['Qtr'] = hp2.apply(lambda x: pd.Period(x['Mth'],'Q'),axis=1)
回答by MaxU
I would use to_datetime()method in a "vectorized" manner:
我会以“矢量化”的方式使用to_datetime()方法：
In [76]: x
Out[76]:
Month
0 2016-11
1 2011-01
2 2015-07
3 2012-09
In [77]: x['Qtr'] = pd.to_datetime(x.Month).dt.quarter
In [78]: x
Out[78]:
Month Qtr
0 2016-11 4
1 2011-01 1
2 2015-07 3
3 2012-09 3
Or if you want to have it in 2016Q4format (as @root mentioned), using PeriodIndex():
或者，如果您想将其2016Q4格式化（如@root 所述），请使用PeriodIndex()：
In [114]: x['Qtr'] = pd.PeriodIndex(pd.to_datetime(x.Mth), freq='Q')
In [115]: x
Out[115]:
Mth Qtr
0 2016-11 2016Q4
1 2011-01 2011Q1
2 2015-07 2015Q3
3 2012-09 2012Q3
回答by neuromusic
Since you don't need the whole row, is it faster if you map the values from the column alone?
由于您不需要整行，如果您单独映射列中的值会更快吗？
hp2['Qtr'] = hp2['Mth'].map(lambda x: pd.Period(x,'Q'))
回答by J. Khoury
I happen to be working on a df that contains 9994 rows so I tested your code against what I've used in the past and posted the results for you. Here is a sample of the df, not exactly YYYY-MM but it doesn't matter because the code will work on either:
我碰巧正在处理一个包含 9994 行的 df，所以我根据我过去使用的内容测试了您的代码，并为您发布了结果。这是 df 的一个示例，不完全是 YYYY-MM，但这并不重要，因为代码可以在任何一个上工作：
hp2['Mth'][:10]
Out[11]:
0 2016-06-26
1 2016-06-26
2 2016-06-26
3 2016-06-26
4 2016-06-26
5 2016-06-26
6 2016-06-26
7 2016-06-26
8 2016-06-26
9 2016-06-26
Name: Mth, dtype: datetime64[ns]
I ran your code on my df:
我在我的 df 上运行了你的代码：
%timeit hp2['Qtr_Period']= hp2.apply(lambda x: pd.Period(x['Mth'],'Q'), axis=1)
hp2['Qtr_Period'][:10]
1 loop, best of 3: 2.28 s per loop
Out[13]:
0 2016Q2
1 2016Q2
2 2016Q2
3 2016Q2
4 2016Q2
5 2016Q2
6 2016Q2
7 2016Q2
8 2016Q2
9 2016Q2
Name: Qtr_Period, dtype: object
Then I tested it using this:
然后我用这个测试了它：
%timeit hp2['Qtr_dt']= (df['Order Date'].dt.year.astype(str))+'Q'+(df['Order Date'].dt.quarter.astype(str))
hp2['Qtr_dt'][:10]
10 loops, best of 3: 67.6 ms per loop
Out[14]:
0 2016Q2
1 2016Q2
2 2016Q2
3 2016Q2
4 2016Q2
5 2016Q2
6 2016Q2
7 2016Q2
8 2016Q2
9 2016Q2
Name: Qtr_dt, dtype: object
It is clear from the results. Hope that helps. You can find more information on pandas.Series.dt
从结果可以看出。希望有帮助。您可以在pandas.Series.dt上找到更多信息
回答by Moses Njenga
month = ['2016-11', '2011-01', '2015-06', '2012-09']
x = pd.DataFrame(month, columns=["month"])
x.month = pd.to_datetime(x.month)
x['quarter'] = [pd.Period(x.month[i], freq='M').quarter for i in range(len(x))]
x
month quarter
0 2016-11-01 4
1 2011-01-01 1
2 2015-06-01 2
3 2012-09-01 3
回答by root
Same idea as @MaxU but using astype:
与@MaxU 相同的想法，但使用astype：
hp2['Qtr'] = pd.to_datetime(hp2['Mth'].values, format='%Y-%m').astype('period[Q]')
The resulting output:
结果输出：
Mth Qtr
0 2014-01 2014Q1
1 2017-02 2017Q1
2 2016-03 2016Q1
3 2017-04 2017Q2
4 2016-05 2016Q2
5 2016-06 2016Q2
6 2017-07 2017Q3
7 2016-08 2016Q3
8 2017-09 2017Q3
9 2015-10 2015Q4
10 2017-11 2017Q4
11 2015-12 2015Q4
Timings
时间安排
Using the following setup to produce a large sample dataset:
使用以下设置生成大型样本数据集：
n = 10**5
yrs = np.random.choice(range(2010, 2021), n)
mths = np.random.choice(range(1, 13), n)
df = pd.DataFrame({'Mth': ['{0}-{1:02d}'.format(*p) for p in zip(yrs, mths)]})
I get the following timings:
我得到以下时间：
%timeit pd.to_datetime(df['Mth'].values, format='%Y-%m').astype('period[Q]')
10 loops, best of 3: 33.4 ms per loop
%timeit pd.PeriodIndex(pd.to_datetime(df.Mth), freq='Q')
1 loop, best of 3: 2.68 s per loop
%timeit df['Mth'].map(lambda x: pd.Period(x,'Q'))
1 loop, best of 3: 6.26 s per loop
%timeit df.apply(lambda x: pd.Period(x['Mth'],'Q'),axis=1)
1 loop, best of 3: 9.49 s per loop
