Question

我遇到了需要帮助的熊猫问题。

一方面，我有一个如下所示的DataFrame：

   contributor_id     timestamp     edits    upper_month   lower_month
0      8             2018-01-01       1      2018-04-01    2018-02-01
1      26424341      2018-01-01       11     2018-04-01    2018-02-01
10     26870381      2018-01-01       465    2018-04-01    2018-02-01
22     28109145      2018-03-01       17     2018-06-01    2018-04-01
23     32769624      2018-01-01       84     2018-04-01    2018-02-01
25     32794352      2018-01-01       4      2018-04-01    2018-02-01

另一方面，我（在另一个DF中可用）具有给定的日期索引：

2018-01-01, 2018-02-01, 2018-03-01, 2018-04-01, 2018-05-01, 2018-06-01, 2018-07-01, 2018-08-01, 2018-09-01, 2018-10-01, 2018-11-01, 2018-12-01.

我需要创建一个pd.Series，它具有先前显示的索引作为索引。对于索引中的每个日期，pd.Series的数据必须为：

如果日期> =较低月份和日期<=较高月份，则我添加1。

目标是按每个日期计算该日期在上一个DataFrame中的上月和下月值之间的次数。

在这种情况下，示例输出pd.Series将为：

2018-01-01    0
2018-02-01    5
2018-03-01    5
2018-04-01    6
2018-05-01    1
2018-06-01    1
2018-07-01    0
2018-08-01    0
2018-09-01    0
2018-10-01    0
2018-11-01    0
2018-12-01    0

是否有一种快速的计算方法，可以避免遍历第一个数据帧很多次？

谢谢大家。

Answer 1

将列表推导和扁平化用于转换为元组和范围值的压缩列之间的测试成员资格，在生成器中创建DataFrame和sum：

rng = pd.date_range('2018-01-01', freq='MS', periods=12)
vals = list(zip(df['lower_month'], df['upper_month']))

s = pd.Series({y: sum(y >= x1 and y <= x2 for x1, x2 in vals) for y in rng})

编辑：

使用count方法以获得更好的性能，谢谢@Stef：

s = pd.Series({y: [y >= x1 and y <= x2 for x1, x2 in vals].count(True) for y in rng})

print (s)
2018-01-01    0
2018-02-01    5
2018-03-01    5
2018-04-01    6
2018-05-01    1
2018-06-01    1
2018-07-01    0
2018-08-01    0
2018-09-01    0
2018-10-01    0
2018-11-01    0
2018-12-01    0
dtype: int64

性能：

np.random.seed(123)

def random_dates(start, end, n=10000):

    start_u = start.value//10**9
    end_u = end.value//10**9

    return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s').floor('d')


d1 = random_dates(pd.to_datetime('2015-01-01'), pd.to_datetime('2018-01-01')) + pd.offsets.MonthBegin(0)
d2 = random_dates(pd.to_datetime('2018-01-01'), pd.to_datetime('2020-01-01')) + pd.offsets.MonthBegin(0)

df = pd.DataFrame({'lower_month':d1, 'upper_month':d2})
rng = pd.date_range('2015-01-01', freq='MS', periods=6 * 12)
vals = list(zip(df['lower_month'], df['upper_month']))

In [238]: %timeit pd.Series({y: [y >= x1 and y <= x2 for x1, x2 in vals].count(True) for y in rng})
158 ms ± 2.55 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [239]: %timeit pd.Series({y: sum(y >= x1 and y <= x2 for x1, x2 in vals) for y in rng})
221 ms ± 17 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

#first solution is slow    
In [240]: %timeit pd.DataFrame([(y, y >= x1 and y <= x2) for x1, x2 in vals for y in rng],                  columns=['d','test']).groupby('d')['test'].sum().astype(int)
4.52 s ± 396 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 2

我已经使用itertools为每个index_date重复上个月和下个月

然后比较每个上个月和下个月的index_date并设置临时列

LocalDate from = yourdate; LocalDate to = yourotherdate; int difference = to.getMonthValue() - from.getMonthValue()) + ((to.getYear() - from.getYear()) * 12) + 1;

然后按索引日期对总和进行检查

check = 1

import pandas as pd
from pandas.compat import StringIO, BytesIO
import itertools

#sample data
data = ('contributor_id,timestamp,edits,upper_month,lower_month\n'
'8,2018-01-01,1,2018-04-01,2018-02-01\n'
'26424341,2018-01-01,11,2018-04-01,2018-02-01\n'
'26870381,2018-02-01,465,2018-04-01,2018-02-01\n'
'28109145,2018-03-01,17,2018-06-01,2018-04-01\n')

orig_df = pd.read_csv(StringIO(data))

# sample index_dates
index_df = list(pd.Series(["2018-01-01", "2018-02-01"]))

# repeat upper_month and lower_month using itertools.product
abc = list(orig_df[['upper_month','lower_month']].values)
combine_list = [index_df,abc]
res = list(itertools.product(*combine_list))
df = pd.DataFrame(res,columns=["timestamp","range"])

#separate lower_month and upper_month from  range 
df['lower_month'] = df['range'].apply(lambda x : x[1])
df['upper_month'] = df['range'].apply(lambda x : x[0])
df.drop(['range'],axis=1,inplace=True)

# convert all dates column to make them consistent
orig_df['timestamp'] = pd.to_datetime(orig_df['timestamp']).dt.date.astype(str)
orig_df['upper_month'] = pd.to_datetime(orig_df['upper_month']).dt.date.astype(str)
orig_df['lower_month'] = pd.to_datetime(orig_df['lower_month']).dt.date.astype(str)

#apply condition to set check 1
df.loc[(df["timestamp"]>=df['lower_month']) & (df["timestamp"]<=df['upper_month']),"check"] = 1

#simply groupby to count check
res = df.groupby(['timestamp'])['check'].sum()

print(res)

计算数据框列的最快方法

2 个答案: