with below example:
df = pd.DataFrame({'signal':[1,0,0,1,0,0,0,0,1,0,0,1,0,0],'product':['A','A','A','A','A','A','A','B','B','B','B','B','B','B'],'price':[1,2,3,4,5,6,7,1,2,3,4,5,6,7],'price2':[1,2,1,2,1,2,1,2,1,2,1,2,1,2]})
I have a function "fill_price" to create a new column 'Price_B' based on 'signal' and 'price'. For every 'product' subgroup, Price_B equals to Price if 'signal' is 1. Price_B equals previous row's Price_B if signal is 0. If the subgroup starts with a 0 'signal', then 'price_B' will be kept at 0 until 'signal' turns 1.
Currently I have:
def fill_price(df, signal,price_A):
p = df[price_A].where(df[signal] == 1)
return p.ffill().fillna(0).astype(df[price_A].dtype)
this is then applied using:
df['Price_B'] = fill_price(df,'signal','price')
However, I want to use df.groupby('product').apply() to apply this fill_price function to two subsets of 'product' columns separately, and also apply it to both'price' and 'price2' columns. Could someone help with that?
I basically want to do:
df.groupby('product',groupby_keys=False).apply(fill_price, 'signal','price2')
IIUC, you can use this syntax:
df['Price_B'] = df.groupby('product').apply(lambda x: fill_price(x,'signal','price2')).reset_index(level=0, drop=True)
Output:
price price2 product signal Price_B
0 1 1 A 1 1
1 2 2 A 0 1
2 3 1 A 0 1
3 4 2 A 1 2
4 5 1 A 0 2
5 6 2 A 0 2
6 7 1 A 0 2
7 1 2 B 0 0
8 2 1 B 1 1
9 3 2 B 0 1
10 4 1 B 0 1
11 5 2 B 1 2
12 6 1 B 0 2
13 7 2 B 0 2
You can write this much simplier without the extra function.
df['Price_B'] = (df.groupby('product',as_index=False)
.apply(lambda x: x['price2'].where(x.signal==1).ffill().fillna(0))
.reset_index(level=0, drop=True))
Related
I would like to drop the [] for a given df
df=pd.DataFrame(dict(a=[1,2,4,[],5]))
Such that the expected output will be
a
0 1
1 2
2 4
3 5
Edit:
or to make thing more interesting, what if we have two columns and some of the cell is with [] to be dropped.
df=pd.DataFrame(dict(a=[1,2,4,[],5],b=[2,[],1,[],6]))
One way is to get the string repr and filter:
df = df[df['a'].map(repr)!='[]']
Output:
a
0 1
1 2
2 4
4 5
For multiple columns, we could apply the above:
out = df[df.apply(lambda c: c.map(repr)).ne('[]').all(axis=1)]
Output:
a b
0 1 2
2 4 1
4 5 6
You can't use equality directly as pandas will try to align a Series and a list, but you can use isin:
df[~df['a'].isin([[]])]
output:
a
0 1
1 2
2 4
4 5
To act on all columns:
df[~df.isin([[]]).any(1)]
output:
a b
0 1 2
2 4 1
4 5 6
This seems to be easy but couldn't find a working solution for it:
I have a dataframe with 3 columns:
df = pd.DataFrame({'A': [0,0,2,2,2],
'B': [1,1,2,2,3],
'C': [1,1,2,3,4]})
A B C
0 0 1 1
1 0 1 1
2 2 2 2
3 2 2 3
4 2 3 4
I want to select rows based on values of column A, then groupby based on values of column B, and finally transform values of column C into sum. something along the line of this (obviously not working) code:
df[df['A'].isin(['2']), 'C'] = df[df['A'].isin(['2']), 'C'].groupby('B').transform('sum')
desired output for above example is:
A B C
0 0 1 1
1 0 1 1
2 2 2 5
3 2 3 4
I also know how to split dataframe and do it. I am looking more for a solution that does it without the need of split+concat/merge. Thank you.
Is it just
s = df['A'].isin([2])
pd.concat((df[s].groupby(['A','B'])['C'].sum().reset_index(),
df[~s])
)
Output:
A B C
0 2 2 5
1 2 3 4
0 0 1 1
Update: Without splitting, you can assign a new column indicating special values of A:
(df.sort_values('A')
.assign(D=(~df['A'].isin([2])).cumsum())
.groupby(['D','A','B'])['C'].sum()
.reset_index('D',drop=True)
.reset_index()
)
Output:
A B C
0 0 1 1
1 0 1 1
2 2 2 5
3 2 3 4
Here is the first DataFrame:
In: df.head()
Out:
avg_lmp avg_load
read_year read_month trading_block
2017 3 0 24.606666 0.018033
1 32.090800 0.023771
4 0 25.136200 0.017487
1 33.487529 0.023570
5 0 24.085170 0.018008
And here is the second DataFrame that I want to join to the first one based on month (even if it has to repeat values such as read_year = 2018 and read_month = 3. If it's 2019 and the read_month is 3, I want it to say the same value for read_month 3.
In: df2.head()
Out:
fpc
read_month trading_block
1 0 37.501837
1 45.750000
2 0 35.531818
1 41.550000
3 0 28.348427
1 35.900000
4 0 26.250870
1 34.150000
5 0 23.599388
1 34.550000
6 0 25.617027
1 38.670000
7 0 27.531765
1 42.050000
8 0 26.628298
1 40.400000
9 0 25.201923
1 36.500000
10 0 25.299149
1 35.250000
11 0 25.349091
1 34.300000
12 0 28.249623
1 35.500000
Is it clear what I'm asking for?
You seem to have common indexes. Set them, then join:
df = df.reset_index().set_index(['read_month', 'trading_block']).join(df2)
and if you wish:
df.reset_index().set_index(['read year', 'read_month', 'trading_block'])
Not sure if that is what you're after:
index avg_lmp avg_load fpc
read_year read_month trading_block
2017 3 0 0 24.606666 0.018033 28.348427
1 1 32.090800 0.023771 35.900000
4 0 2 25.136200 0.017487 26.250870
1 3 33.487529 0.023570 34.150000
5 0 4 24.085170 0.018008 23.599388
maybe try this. just merge on the two columns with a outer. (like full outer)
DataFrame1.merge(DataFrame2, left_on='read_month', right_on='read_month', how='outer')
Consider the following dataframe:
>>> import pandas as pd
>>> df = pd.DataFrame({'group': list('aaabbabc')})
>>> df
group
0 a
1 a
2 a
3 b
4 b
5 a
6 b
7 c
I want to count the cumulative number of times each group has occurred. My desired output looks like this:
>>> df
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
My initial approach was to do something like this:
df['n'] = df.groupby('group').apply(lambda x: list(range(x.shape[0])))
Basically assigning a length n array, zero-indexed, to each group. But that has proven difficult to transpose and join.
You can use groupby + cumcount, and horizontally concat the new column:
>>> pd.concat([df, df.group.groupby(df.group).cumcount()], axis=1).rename(columns={0: 'n'})
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
Simply use groupby on column name, in this case group and then apply cumcount and finally add a column in dataframe with the result.
df['n']=df.groupby('group').cumcount()
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
You can use apply method by passing a lambda expression as parameter.
The idea is that you need to find out the count for a group as number of appearances for that group from the previous rows.
df['n'] = df.apply(lambda x: list(df['group'])[:int(x.name)].count(x['group']), axis=1)
Output
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
Note: cumcount method is build with the help of the apply function.
You can read this in pandas documentation.
I don't have much experience with working with pandas. I have a pandas dataframe as shown below.
df = pd.DataFrame({ 'A' : [1,2,1],
'start' : [1,3,4],
'stop' : [3,4,8]})
I would like to create a new dataframe that iterates through the rows and appends to resulting dataframe. For example, from row 1 of the input dataframe - Generate a sequence of numbers [1,2,3] and corresponding column to named 1
A seq
1 1
1 2
1 3
2 3
2 4
1 4
1 5
1 6
1 7
1 8
So far, I've managed to identify what function to use to iterate through the rows of the pandas dataframe.
Here's one way with apply:
(df.set_index('A')
.apply(lambda x: pd.Series(np.arange(x['start'], x['stop'] + 1)), axis=1)
.stack()
.to_frame('seq')
.reset_index(level=1, drop=True)
.astype('int')
)
Out:
seq
A
1 1
1 2
1 3
2 3
2 4
1 4
1 5
1 6
1 7
1 8
If you would want to use loops.
In [1164]: data = []
In [1165]: for _, x in df.iterrows():
...: data += [[x.A, y] for y in range(x.start, x.stop+1)]
...:
In [1166]: pd.DataFrame(data, columns=['A', 'seq'])
Out[1166]:
A seq
0 1 1
1 1 2
2 1 3
3 2 3
4 2 4
5 1 4
6 1 5
7 1 6
8 1 7
9 1 8
To add to the answers above, here's a method that defines a function for interpreting the dataframe input shown, into a form that the poster wants:
def gen_df_permutations(perm_def_df):
m_list = []
for i in perm_def_df.index:
row = perm_def_df.loc[i]
for n in range(row.start, row.stop+1):
r_list = [row.A,n]
m_list.append(r_list)
return m_list
Call it, referencing the specification dataframe:
gen_df_permutations(df)
Or optionally call it wrapped in a dataframe creation function to return a final dataframe output:
pd.DataFrame(gen_df_permutations(df),columns=['A','seq'])
A seq
0 1 1
1 1 2
2 1 3
3 2 3
4 2 4
5 1 4
6 1 5
7 1 6
8 1 7
9 1 8
N.B. the first column there is the dataframe index that can be removed/ignored as requirements allow.