I have a time series data given below:
date product price amount
11/01/2019 A 10 20
11/02/2019 A 10 20
11/03/2019 A 25 15
11/04/2019 C 40 50
11/05/2019 C 50 60
I have a high dimensional data, and I have just added the simplified version with two columns {price, amount}. I am trying to transform it relatively based on time index illustrated below:
date product price amount
11/01/2019 A NaN NaN
11/02/2019 A 0 0
11/03/2019 A 15 -5
11/04/2019 C NaN NaN
11/05/2019 C 10 10
I am trying to get relative changes of each product based on time indexes. If previous date does not exist for a specified product, I am adding "NaN".
Can you please tell me is there any function to do this?
Group by product and use .diff()
df[["price", "amount"]] = df.groupby("product")[["price", "amount"]].diff()
output :
date product price amount
0 2019-11-01 A NaN NaN
1 2019-11-02 A 0.0 0.0
2 2019-11-03 A 15.0 -5.0
3 2019-11-04 C NaN NaN
4 2019-11-05 C 10.0 10.0
Related
I have a dataframe of the following scheme in pyspark:
user_id datadate page_1.A page_1.B page_1.C page_2.A page_2.B \
0 111 20220203 NaN NaN NaN NaN NaN
1 222 20220203 5 5 5 5.0 5.0
2 333 20220203 3 3 3 3.0 3.0
page_2.C page_3.A page_3.B page_3.C
0 NaN 1.0 1.0 2.0
1 5.0 NaN NaN NaN
2 4.0 NaN NaN NaN
So it contains columns like user_id, datadate, and few columns for each page (got 3 pages), which are the result of 2 joins. In this example, i have page_1, page_2, page_3, and each has 3 columns: A,B,C. Additionally, for each page columns, for each row, they will either be all null or all full, like in my example.
I don't care about the values of each of the columns per page, I just want to get for each row, the [A,B,C] values that are not null.
example for a wanted result table:
user_id datadate A B C
0 111 20220203 1 1 2
1 222 20220203 5 5 5
2 333 20220203 3 3 3
so the logic will be something like:
df[A] = page_1.A or page_2.A or page_3.A, whichever is not null
df[B] = page_1.B or page_2.B or page_3.B, whichever is not null
df[C] = page_1.C or page_2.C or page_3.C, whichever is not null
for all of the rows..
and of course, I would like to do it in an efficient way.
Thanks a lot.
You can use the sql functions greatest to extract the greatest values in a list of columns.
You can find the documentation here: https://spark.apache.org/docs/3.1.1/api/python/reference/api/pyspark.sql.functions.greatest.html
from pyspark.sql import functions as F
(df.withColumn('A', F.greates(F.col('page_1.A'), F.col('page_2.A), F.col('page_3.A'))
.withColumn('B', F.greates(F.col('page_1.B'), F.col('page_2.B), F.col('page_3.B'))
.select('userid', 'datadate', 'A', 'B'))
This question already has answers here:
Display rows with one or more NaN values in pandas dataframe
(5 answers)
Closed 2 years ago.
I was validating 'Price' column in my dataframe. Sample:
ArticleId SiteId ZoneId Date Quantity Price CostPrice
53 194516 9 2 2018-11-26 11.0 40.64 27.73
164 200838 9 2 2018-11-13 5.0 99.75 87.24
373 200838 9 2 2018-11-27 1.0 99.75 87.34
pd.to_numeric(df_sales['Price'], errors='coerce').notna().value_counts()
And I'd love to display those rows with False values so I know whats wrong with them. How do I do that?
True 17984
False 13
Name: Price, dtype: int64
Thank you.
You could print your rows when price isnull():
print(df_sales[df_sales['Price'].isnull()])
ArticleId SiteId ZoneId Date Quantity Price CostPrice
1 200838 9 2 2018-11-13 5 NaN 87.240
pd.to_numeric(df['Price'], errors='coerce').isna() returns a Boolean, which can be used to select the rows that cause errors.
This includes NaN or rows with strings
import pandas as pd
# test data
df = pd.DataFrame({'Price': ['40.64', '99.75', '99.75', pd.NA, 'test', '99. 0', '98 0']})
Price
0 40.64
1 99.75
2 99.75
3 <NA>
4 test
5 99. 0
6 98 0
# find the value of the rows that are causing issues
problem_rows = df[pd.to_numeric(df['Price'], errors='coerce').isna()]
# display(problem_rows)
Price
3 <NA>
4 test
5 99. 0
6 98 0
Alternative
Create an extra column and then use it to select the problem rows
df['Price_Updated'] = pd.to_numeric(df['Price'], errors='coerce')
Price Price_Updated
0 40.64 40.64
1 99.75 99.75
2 99.75 99.75
3 <NA> NaN
4 test NaN
5 99. 0 NaN
6 98 0 NaN
# find the problem rows
problem_rows = df.Price[df.Price_Updated.isna()]
Explanation
Updating the column with .to_numeric(), and then checking for NaNs will not tell you why the rows had to be coerced.
# update the Price row
df.Price = pd.to_numeric(df['Price'], errors='coerce')
# check for NaN
problem_rows = df.Price[df.Price.isnull()]
# display(problem_rows)
3 NaN
4 NaN
5 NaN
6 NaN
Name: Price, dtype: float64
I have a dataset given below:
weekid type amount
1 A 10
1 B 20
1 C 30
1 D 40
1 F 50
2 A 70
2 E 80
2 B 100
I am trying to convert it to another panda frame based on total number of type values defined with:
import pandas as pd
import numpy as np
df=pd.read_csv(INPUT_FILE)
for type in df["type"].unique():
//todo
My aim is to get a data given below:
weekid type_A type_B type_C type_D type_E type_F
1 10 20 30 40 0 50
2 70 100 0 0 80 0
Is there any specific function that convert unique values as a column and fills the missing values as 0 for each weekId groups? I am wondering that how this conversion can be done efficiently?
You can use the following:
df = df.pivot(columns=['type'], values=['amount'])
df.fillna(0)
dfp.columns = dfp.columns.droplevel(0)
Given your input this yields:
type A B C D F
weekid
1 10.0 20.0 30.0 40.0 50.0
2 70.0 80.0 100.0 0.0 0.0
I have a dataframe which contains timeseries data. What i want to do is efficiently fill all the missing values in different columns by substituting with a median value using timedelta of say "N" mins. E.g if for a column say i have data for 10:20, 10:21,10:22,10:23,10:24,.... and data in 10:22 is missing then with timedelta of say 2 mins i would want it to be filled by median value of 10:20,10:21,10:23 and 10:24.
One way i can do is :
for all column in dataframe:
Find index which has nan value
for all index which has nan value:
extract all values using between_time with index-timedelta and index_+deltatime
find the media of extracted value
set value in the index with that extracted median value.
This looks like 2 for loops running and not a very efficient one. Is there a efficient way to do it.
Thanks
IIUC you can resample your time column, then fillna with rolling window set to center:
# dummy data setup
np.random.seed(500)
n = 2
df = pd.DataFrame({"time":pd.to_timedelta([f"10:{i}:00" for i in range(15)]),
"value":np.random.randint(2, 10, 15)})
df = df.drop(df.index[[5,10]]).reset_index(drop=True)
print (df)
time value
0 10:00:00 4
1 10:01:00 9
2 10:02:00 3
3 10:03:00 3
4 10:04:00 8
5 10:06:00 9
6 10:07:00 2
7 10:08:00 9
8 10:09:00 9
9 10:11:00 7
10 10:12:00 3
11 10:13:00 3
12 10:14:00 7
s = df.set_index("time").resample("60S").asfreq()
print (s.fillna(s.rolling(n*2+1, min_periods=1, center=True).mean()))
value
time
10:00:00 4.0
10:01:00 9.0
10:02:00 3.0
10:03:00 3.0
10:04:00 8.0
10:05:00 5.5
10:06:00 9.0
10:07:00 2.0
10:08:00 9.0
10:09:00 9.0
10:10:00 7.0
10:11:00 7.0
10:12:00 3.0
10:13:00 3.0
10:14:00 7.0
I want to sort and group by a pandas data frame column alphabetically.
a b c
0 sales 2 NaN
1 purchase 130 230.0
2 purchase 10 20.0
3 sales 122 245.0
4 purchase 103 320.0
I want to sort column "a" such that it is in alphabetical order and is grouped as well i.e., the output is as follows:
a b c
1 purchase 130 230.0
2 10 20.0
4 103 320.0
0 sales 2 NaN
3 122 245.0
How can I do this?
I think you should use the sort_values method of pandas :
result = dataframe.sort_values('a')
It will sort your dataframe by the column a and it will be grouped either because of the sorting. See ya !