I'm trying to create a new column in a DataFrame and storing it with values stored in a different dataframe by first comparing the values of columns that both dataframes have. For example:
df1 >>>
| name | team | week | dates | interceptions | pass_yds | rating |
| ---- | ---- | -----| ---------- | ------------- | --------- | -------- |
| maho | KC | 1 | 2020-09-10 | 0 | 300 | 105 |
| went | PHI | 1 | 2020-09-13 | 2 | 225 | 74 |
| lock | DEN | 1 | 2020-09-14 | 0 | 150 | 89 |
| dris | DEN | 2 | 2020-09-20 | 1 | 220 | 95 |
| went | PHI | 2 | 2020-09-20 | 2 | 250 | 64 |
| maho | KC | 2 | 2020-09-21 | 1 | 245 | 101 |
df2 >>>
| name | team | week | catches | rec_yds | rec_tds |
| ---- | ---- | -----| ------- | ------- | ------- |
| ertz | PHI | 1 | 5 | 58 | 1 |
| fant | DEN | 2 | 6 | 79 | 0 |
| kelc | KC | 2 | 8 | 105 | 1 |
| fant | DEN | 1 | 3 | 29 | 0 |
| kelc | KC | 1 | 6 | 71 | 1 |
| ertz | PHI | 2 | 7 | 91 | 2 |
| goed | PHI | 2 | 2 | 15 | 0 |
I want to create a dates column in df2 with the values of the dates stored in the dates column in df1 after matching the teams and the weeks columns. After the matching, df2 in this example should look something like this:
df2 >>>
| name | team | week | catches | rec_yds | rec_tds | dates |
| ---- | ---- | -----| ------- | ------- | ------- | ---------- |
| ertz | PHI | 1 | 5 | 58 | 1 | 2020-09-13 |
| fant | DEN | 2 | 6 | 79 | 0 | 2020-09-20 |
| kelc | KC | 2 | 8 | 105 | 1 | 2020-09-20 |
| fant | DEN | 1 | 3 | 29 | 0 | 2020-09-14 |
| kelc | KC | 1 | 6 | 71 | 1 | 2020-09-10 |
| ertz | PHI | 2 | 7 | 91 | 2 | 2020-09-20 |
| goed | PHI | 2 | 2 | 15 | 0 | 2020-09-20 |
I'm looking for an optimal solution. I've already tried nested for loops and comparing the week and team columns from both dataframes together but that hasn't worked. At this point I'm all out of ideas. Please help!
Disclaimer: The actual DataFrames I'm working with are a lot larger. They have a lot more rows, columns, and values (i.e. a lot more teams in the team columns, a lot more dates in the dates columns, and a lot more weeks in the week columns)
Related
I was asking the question in this link SUMIFS in python jupyter
However, I just realized that the solution didn't work because they can switch in and switch out on different dates. So basically they have to switch out first before they can switch in.
Here is the dataframe (sorted based on the date):
+---------------+--------+---------+-----------+--------+
| Switch In/Out | Client | Quality | Date | Amount |
+---------------+--------+---------+-----------+--------+
| Out | 1 | B | 15-Aug-19 | 360 |
| In | 1 | A | 16-Aug-19 | 180 |
| In | 1 | B | 17-Aug-19 | 180 |
| Out | 1 | A | 18-Aug-19 | 140 |
| In | 1 | B | 18-Aug-19 | 80 |
| In | 1 | A | 19-Aug-19 | 60 |
| Out | 2 | B | 14-Aug-19 | 45 |
| Out | 2 | C | 15-Aug-20 | 85 |
| In | 2 | C | 15-Aug-20 | 130 |
| Out | 2 | A | 20-Aug-19 | 100 |
| In | 2 | A | 22-Aug-19 | 30 |
| In | 2 | B | 23-Aug-19 | 30 |
| In | 2 | C | 23-Aug-19 | 40 |
+---------------+--------+---------+-----------+--------+
I would then create a new column and divide them into different transactions.
+---------------+--------+---------+-----------+--------+------+
| Switch In/Out | Client | Quality | Date | Amount | Rows |
+---------------+--------+---------+-----------+--------+------+
| Out | 1 | B | 15-Aug-19 | 360 | 1 |
| In | 1 | A | 16-Aug-19 | 180 | 1 |
| In | 1 | B | 17-Aug-19 | 180 | 1 |
| Out | 1 | A | 18-Aug-19 | 140 | 2 |
| In | 1 | B | 18-Aug-19 | 80 | 2 |
| In | 1 | A | 19-Aug-19 | 60 | 2 |
| Out | 2 | B | 14-Aug-19 | 45 | 3 |
| Out | 2 | C | 15-Aug-20 | 85 | 3 |
| In | 2 | C | 15-Aug-20 | 130 | 3 |
| Out | 2 | A | 20-Aug-19 | 100 | 4 |
| In | 2 | A | 22-Aug-19 | 30 | 4 |
| In | 2 | B | 23-Aug-19 | 30 | 4 |
| In | 2 | C | 23-Aug-19 | 40 | 4 |
+---------------+--------+---------+-----------+--------+------+
With this, I can apply the pivot formula and take it from there.
However, how do I do this in python? In excel, I can just use multiple SUMIFS and compare in and out. However, this is not possible in python.
Thank you!
One simple solution is to iterate and apply a check (function) over each element being the result a new column, so: map.
Using df.index.map we get the index for each item to pass as a argument, so we can play with the values, get and compare. In your case your aim is to identify the change to "Out" keeping a counter.
import pandas as pd
switchInOut = ["Out", "In", "In", "Out", "In", "In",
"Out", "Out", "In", "Out", "In", "In", "In"]
df = pd.DataFrame(switchInOut, columns=['Switch In/Out'])
counter = 1
def changeToOut(i):
global counter
if df["Switch In/Out"].get(i) == "Out" and df["Switch In/Out"].get(i-1) == "In":
counter += 1
return counter
rows = df.index.map(changeToOut)
df["Rows"] = rows
df
Result:
+----+-----------------+--------+
| | Switch In/Out | Rows |
|----+-----------------+--------|
| 0 | Out | 1 |
| 1 | In | 1 |
| 2 | In | 1 |
| 3 | Out | 2 |
| 4 | In | 2 |
| 5 | In | 2 |
| 6 | Out | 3 |
| 7 | Out | 3 |
| 8 | In | 3 |
| 9 | Out | 4 |
| 10 | In | 4 |
| 11 | In | 4 |
| 12 | In | 4 |
+----+-----------------+--------+
I have a dataframe like below
+-----------+------------+---------------+------+-----+-------+
| InvoiceNo | CategoryNo | Invoice Value | Item | Qty | Price |
+-----------+------------+---------------+------+-----+-------+
| 1 | 1 | 77 | 128 | 1 | 10 |
| 1 | 1 | 77 | 101 | 1 | 11 |
| 1 | 2 | 77 | 105 | 3 | 12 |
| 1 | 3 | 77 | 129 | 2 | 10 |
| 2 | 1 | 21 | 145 | 1 | 9 |
| 2 | 2 | 21 | 130 | 1 | 12 |
+-----------+------------+---------------+------+-----+-------+
I want to filter the entire group, if any of the items in the list item_list = [128,129,130] is present in that group, after grouping by 'InvoiceNo' &'CategoryNo'.
My desired out put is as below
+-----------+------------+---------------+------+-----+-------+
| InvoiceNo | CategoryNo | Invoice Value | Item | Qty | Price |
+-----------+------------+---------------+------+-----+-------+
| 1 | 1 | 77 | 128 | 1 | 10 |
| 1 | 1 | 77 | 101 | 1 | 11 |
| 1 | 3 | 77 | 129 | 2 | 10 |
| 2 | 2 | 21 | 130 | 1 | 12 |
+-----------+------------+---------------+------+-----+-------+
I know how to filter a dataframe using isin(). But, not sure how to do it with groupby()
so far i have tried below
import pandas as pd
df = pd.read_csv('data.csv')
item_list = [128,129,130]
df.groupby(['InvoiceNo','CategoryNo'])['Item'].isin(item_list)
but nothing happens. please guide me how to solve this issue.
You can do something like this:
s = (df['Item'].isin(item_list)
.groupby([df['InvoiceNo'], df['CategoryNo']])
.transform('any')
)
df[s]
Please see this SO post Manipulating pandas columns
I shared this dataframe:
+----------+------------+-------+-----+------+
| Location | Date | Event | Key | Time |
+----------+------------+-------+-----+------+
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-04 | 1 | a | 2 |
| i2 | 2019-03-15 | 2 | b | 0 |
| i9 | 2019-02-22 | 2 | c | 0 |
| i9 | 2019-03-10 | 3 | d | |
| i9 | 2019-03-10 | 3 | d | 0 |
| s8 | 2019-04-22 | 1 | e | |
| s8 | 2019-04-25 | 1 | e | |
| s8 | 2019-04-28 | 1 | e | 6 |
| t14 | 2019-05-13 | 3 | f | |
+----------+------------+-------+-----+------+
This is a follow-up question. Consider two more columns after Date as shown below.
+-----------------------+----------------------+
| Start Time (hh:mm:ss) | Stop Time (hh:mm:ss) |
+-----------------------+----------------------+
| 13:24:38 | 14:17:39 |
| 03:48:36 | 04:17:20 |
| 04:55:05 | 05:23:48 |
| 08:44:34 | 09:13:15 |
| 19:21:05 | 20:18:57 |
| 21:05:06 | 22:01:50 |
| 14:24:43 | 14:59:37 |
| 07:57:32 | 09:46:21
| 19:21:05 | 20:18:57 |
| 21:05:06 | 22:01:50 |
| 14:24:43 | 14:59:37 |
| 07:57:32 | 09:46:21 |
+-----------------------+----------------------+
The task remains the same - to get the time difference but in hours, corresponding to the Stop Time of the first row and Start Time of the last row
for each Key.
Based on the answer, I was trying something like this:
df['Time']=df.groupby(['Location','Event']).Date.\
transform(lambda x : (x.iloc[-1]-x.iloc[0]))[~df.duplicated(['Location','Event'],keep='last')]
df['Time_h']=df.groupby(['Location','Event'])['Start Time (hh:mm:ss)','Stop Time (hh:mm:ss)'].\
transform(lambda x,y : (x.iloc[-1]-y.iloc[0]))[~df.duplicated(['Location','Event'],keep='last')] # This gives an error on transform
to get the difference in days and hours separately and then combine. Is there a better way?
This question already has answers here:
How do I Pandas group-by to get sum?
(11 answers)
Closed 3 years ago.
I have some data like this:
+-----------+---------+-------+
| Duration | Outcome | Event |
+-----------+---------+-------+
| 421 | 0 | 1 |
| 421 | 0 | 1 |
| 261 | 0 | 1 |
| 24 | 0 | 1 |
| 27 | 0 | 1 |
| 613 | 0 | 1 |
| 2454 | 0 | 1 |
| 227 | 0 | 1 |
| 2560 | 0 | 1 |
| 229 | 0 | 1 |
| 2242 | 0 | 1 |
| 6680 | 0 | 1 |
| 1172 | 0 | 1 |
| 5656 | 0 | 1 |
| 5082 | 0 | 1 |
| 7239 | 0 | 1 |
| 127 | 0 | 1 |
| 128 | 0 | 1 |
| 128 | 0 | 1 |
| 7569 | 1 | 1 |
| 324 | 0 | 2 |
| 6395 | 0 | 2 |
| 6196 | 0 | 2 |
| 31 | 0 | 2 |
| 228 | 0 | 2 |
| 274 | 0 | 2 |
| 270 | 0 | 2 |
| 275 | 0 | 2 |
| 232 | 0 | 2 |
| 7310 | 0 | 2 |
| 7644 | 1 | 2 |
| 6949 | 0 | 3 |
| 6903 | 1 | 3 |
| 6942 | 0 | 4 |
| 7031 | 1 | 4 |
+-----------+---------+-------+
Now, for each Event, with the Outcome 0/1 considered as Fail/Pass, I want to sum the total Duration of Fail/Pass events separately in 2 new columns (or 1, whatever ensures readability).
I'm new to dataframes and I feel significant logical indexing is involved here. What is the best way to approach this problem?
df.groupby(['Event', 'Outcome'])['Duration'].sum()
So you group by both the event then the outcome, look at the duration column then take the sum of each group.
You can also try:
pd.pivot_table(index='Event',
columns='Outcome',
values='Duration',
data=df,
aggfunc='sum')
which gives you a table with two columns:
+---------+-------+------+
| Outcome | 0 | 1 |
+---------+-------+------+
| Event | | |
+---------+-------+------+
| 1 | 35691 | 7569 |
| 2 | 21535 | 7644 |
| 3 | 6949 | 6903 |
| 4 | 6942 | 7031 |
+---------+-------+------+
I have a dataframe in a similar format to this:
+--------+--------+----------+------+------+------+------+
| | | | | day1 | day2 | day3 |
+--------+--------+----------+------+------+------+------+
| id_one | id_two | id_three | date | | | |
| 18273 | 50 | 1 | 3 | 9 | 11 | 3 |
| | | | 4 | 26 | 27 | 68 |
| | | | 5 | 92 | 25 | 4 |
| | | | 6 | 60 | 72 | 83 |
| | 60 | 2 | 5 | 69 | 93 | 84 |
| | | | 6 | 69 | 30 | 12 |
| | | | 7 | 65 | 65 | 59 |
| | | | 8 | 57 | 88 | 59 |
| | 70 | 3 | 5 | 22 | 95 | 7 |
| | | | 6 | 40 | 24 | 20 |
| | | | 7 | 73 | 81 | 57 |
| | | | 8 | 43 | 8 | 66 |
+--------+--------+----------+------+------+------+------+
I am trying to create tuple that contains id_one, id_two and the values that each grouping contains.
To test this, I am simply trying to print the ids and values like this:
for id_two, data in df.head(100).groupby(level='id_two'):
print id_two, data.values.ravel()
Which gives me the id_two and the data exactly as it should.
I am running into problems when I try and incorporate id_one. I tried this, but was met with an error ValueError: need more than 2 values to unpack
for id_one, id_two, data in df.head(100).groupby(level='id_two'):
print id_one, id_two, data.values.ravel()
How can I print id_one, id_two and the data?
You can pass a list of columns into the level parameter:
df.head.groupby(level=['id_one', 'id_two'])