I have an array of dataframes dfs = [df0, df1, ...]. Each one of them have a date column of varying size (some dates might be in one dataframe but not the other).
What I'm trying to do is this:
pd.concat(dfs).groupby("date", as_index=False).sum()
But with date no longer being a column but an index (dfs = [df.set_index("date") for df in dfs]).
I've seen you can pass df.index to groupby (.groupby(df.index)) but df.index might not include all the dates.
How can I do this?
The goal here is to call .sum() on the groupby, so I'm not tied to using groupby nor concat is there's any alternative method to do so.
If I am able to understand maybe you want something like this:
df = pd.concat([dfs])
df.groupby(df.index).sum()
Here's small example:
tmp1 = pd.DataFrame({'date':['2019-09-01','2019-09-02','2019-09-03'],'value':[1,1,1]}).set_index('date')
tmp2 = pd.DataFrame({'date':['2019-09-01','2019-09-02','2019-09-04','2019-09-05'],'value':[2,2,2,2]}).set_index('date')
df = pd.concat([tmp1,tmp2])
df.groupby(df.index).sum()
Related
Have the following multi index data frame, df.
I performed a 20 day moving average operation on the df[‘Close’] with the following code, ma20.
How do I append m20 to df as a multi index data frame?
I should have 3 level 0 columns, ['Adj Close','Close', ‘ma20’], each with the 3 tickers, ['MSFT','AAPL','AMZN'], at level 1 columns.
The answer should also not require me to type out all the tickers manually.
import yfinance as yf
df = yf.download(['MSFT','AAPL','AMZN'], start="2022-01-01", end="2022-09-01").loc[:,['Adj Close','Close']]
ma20 = df['Close'].sort_index(ascending=True).rolling(20, min_periods=20).mean()
pd.concat([df,ma20], axis=1)????
It's not very elegant solution, but should work, the idea is to specify explicitly the multiindex names for new columns:
df[[('Mean', col) for col in ma20.columns]] = ma20
UPDATE:
If you want to use concat() method, you need firstly to add another column index level to ma20. The way you can do this looks counter-intuitive up to me:
pd.concat((df, pd.concat({'Mean': ma20}, axis=1)), sort=False, axis=1)
The purpose of pd.concat({'Mean': ma20}, axis=1)) is just to add another index level to the columns of ma20
I have a pandas dataframe with the following data: (in csv)
#list1
poke_id,symbol
0,BTC
1,ETB
2,USDC
#list2
5,SOL
6,XRP
I am able to concatenate them into one dataframe using the following code:
df = pd.concat([df1, df2], ignore_index = True)
df = df.reset_index(drop = True)
df['poke_id'] = df.index
df = df[['poke_id','symbol']]
which gives me the output: (in csv)
poke_id,symbol
0,BTC
1,ETB
2,USDC
3,SOL
4,XRP
Is there any other way to do the same. I think calling the whole data frame of ~4000 entries just to add ~100 more will be a little pointless and cumbersome. How can I make it in such a way that it picks list 1 (or dataframe 1) and picks the highest poke_id; and just does i + 1 to the later entries in list 2.
Your solution is good, is possible simplify:
df = pd.concat([df1, df2], ignore_index = True).rename_axis('poke_id').reset_index()
use indexes to get what data you want from the dataframe, although this is not effective if you want large amounts of data from the dataframe, this method allows you to take specific amounts of data from the dataframe
my df columns names are dates in this format: dd-mm-yy. when I use sort_index(axis = 1) it sort by the first two digits (which specify the days) so it doesn't make sense chronologically. How can I sort it automatically by taking into account also the months?
my df headers:
submitted_at 06-05-18 13-05-18 29-04-18
I expected the output of:
submitted_at 29-04-18 06-05-18 13-05-18
Convert the columns to datetime and use argsort to find the correct ordering. This will put all non-dates to the left in the order they occur, followed by the sorted dates.
import pandas as pd
df = pd.DataFrame(columns=['submitted_at', '06-05-18', '13-05-18', '29-04-18'])
idx = pd.to_datetime(df.columns, errors='coerce', format='%d-%m-%y').argsort()
df.iloc[:, idx]
Empty DataFrame
Columns: [submitted_at, 29-04-18, 06-05-18, 13-05-18]
Converting strings to datetime then sorting them with something like this :
from datetime import datetime
cols_as_date = [datetime.strptime(x,'%d-%m-%Y') for x in df.columns]
df = df[sorted(cols_as_data)]
just convert to DateTime your column
df['newdate']=pd.to_datetime(df.date,format='%d-%m-%y')
and then sort it using sort_values
df.sort_values(by='newdate')
I have 2 dataframes. df1 comprises a Series of values.
df1 = pd.DataFrame({'winnings': cumsums_winnings_s, 'returns':cumsums_returns_s, 'spent': cumsums_spent_s, 'runs': cumsums_runs_s, 'wins': cumsums_wins_s, 'expected': cumsums_expected_s}, columns=["winnings", "returns", "runs", "wins", "expected"])
df2 runs each row through a function which takes 3 columns and produces a result for each row - specialSauce
df2= pd.DataFrame(list(map(lambda w,r,e: doStuff(w,r,e), df1['wins'], df1['runs'], df1['expected'])), columns=["specialSauce"])
print(df2.append(df1))
produces all the df1 columns but NaN for the df1 (and vice versa if df1/df2 switched in append)
So the problem I has is how to append these 2 dataframes correctly.
As I understand things, your issue seems to be related to the fact that you get NaN's in the result DataFrame.
The reason for this is that you are trying to .append() one dataframe to the other while they don't have the same columns.
df2 has one extra column, the one created with apply() and doStuff, while df1 does not have that column. When trying to append one pd.DataFrame to the other the result will have all columns both pd.DataFrame objects. Naturally, you will have some NaN's for ['specialSauce'] since this column does not exist in df1.
This would be the same if you were to use pd.concat(), both methods do the same thing in this case. The one thing that you could do to bring the result closer to your desired result is use the ignore_index flag like this:
>> df2.append(df1, ignore_index=True)
This would at least give you a 'fresh' index for the result pd.DataFrame.
EDIT
If what you're looking for is to "append" the result of doStuff to the end of your existing df, in the form of a new column (['specialSauce']), then what you'll have to do is use pd.concat() like this:
>> pd.concat([df1, df2], axis=1)
This will return the result pd.DataFrame as you want it.
If you had a pd.Series to add to the columns of df1 then you'd need to add it like this:
>> df1['specialSauce'] = <'specialSauce values'>
I hope that helps, if not please rephrase the description of what you're after.
Ok, there are a couple of things going on here. You've left code out and I had to fill in the gaps. For example you did not define doStuff, so I had to.
doStuff = lambda w, r, e: w + r + e
With that defined, your code does not run. I had to guess what you were trying to do. I'm guessing that you want to have an additional column called 'specialSauce' adjacent to your other columns.
So, this is how I set it up and solved the problem.
Setup and Solution
import pandas as pd
import numpy as np
np.random.seed(314)
df = pd.DataFrame(np.random.randn(100, 6),
columns=["winnings", "returns",
"spent", "runs",
"wins", "expected"]).cumsum()
doStuff = lambda w, r, e: w + r + e
df['specialSauce'] = df[['wins', 'runs', 'expected']].apply(lambda x: doStuff(*x), axis=1)
print df.head()
winnings returns spent runs wins expected specialSauce
0 0.166085 0.781964 0.852285 -0.707071 -0.931657 0.886661 -0.752067
1 -0.055704 1.163688 0.079710 0.155916 -1.212917 -0.045265 -1.102266
2 -0.554241 1.928014 0.271214 -0.462848 0.452802 1.692924 1.682878
3 0.627985 3.047389 -1.594841 -1.099262 -0.308115 4.356977 2.949601
4 0.796156 3.228755 -0.273482 -0.661442 -0.111355 2.827409 2.054611
Also
You tried to use pd.DataFrame.append(). Per the linked documentation, it attaches the DataFrame specified as the argument to the end of the DataFrame that is being appended to. You would have wanted to use pd.DataFrame.concat().
I currently have a pandas Series with dtype Timestamp, and I want to group it by date (and have many rows with different times in each group).
The seemingly obvious way of doing this would be something similar to
grouped = s.groupby(lambda x: x.date())
However, pandas' groupby groups Series by its index. How can I make it group by value instead?
grouped = s.groupby(s)
Or:
grouped = s.groupby(lambda x: s[x])
Three methods:
DataFrame: pd.groupby(['column']).size()
Series： sel.groupby(sel).size()
Series to DataFrame:
pd.DataFrame( sel, columns=['column']).groupby(['column']).size()
For anyone else who wants to do this inline without throwing a lambda in (which tends to kill performance):
s.to_frame(0).groupby(0)[0]
You should convert it to a DataFrame, then add a column that is the date(). You can do groupby on the DataFrame with the date column.
df = pandas.DataFrame(s, columns=["datetime"])
df["date"] = df["datetime"].apply(lambda x: x.date())
df.groupby("date")
Then "date" becomes your index. You have to do it this way because the final grouped object needs an index so you can do things like select a group.
To add another suggestion, I often use the following as it uses simple logic:
pd.Series(index=s.values).groupby(level=0)