Develop Reference

Rolling and Mode function to get the majority of voting for rows in pandas Dataframe

I have a pandas Dataframe:
np.random.seed(0)
df = pd.DataFrame({'Close': np.random.uniform(0, 100, size=10)})
lbound, ubound = 0, 1
change = df["Close"].diff()
df["Change"] = change
df["Result"] = np.select([ np.isclose(change, 1) | np.isclose(change, 0) | np.isclose(change, -1),
# The other conditions
(change > 0) & (change > ubound),
(change < 0) & (change < lbound),
change.between(lbound, ubound)],[0, 1, -1, 0])
Close Change Result
0 54.881350 NaN 0
1 71.518937 16.637586 1
2 60.276338 -11.242599 -1
3 54.488318 -5.788019 -1
4 42.365480 -12.122838 -1
5 64.589411 22.223931 1
6 43.758721 -20.830690 -1
7 89.177300 45.418579 1
8 96.366276 7.188976 1
9 38.344152 58.022124 -1
Problem statement - Now, I want the majority of voting for index 1,2,3,4 assigned to index 0, index 2,3,4,5 assigned to index 1 of result columns, and so on for all the subsequent indexes.
I tried:
df['Voting'] = df['Result'].rolling(window = 4,min_periods=1).apply(lambda x: x.mode()[0]).shift()
But,this doesn't give the result I intend. It takes the first 4 rolling window and applies the mode function.
Close Change Result Voting
0 54.881350 NaN 0 NaN
1 71.518937 16.637586 1 0.0
2 60.276338 -11.242599 -1 0.0
3 54.488318 -5.788019 -1 -1.0
4 42.36548 -12.122838 -1 -1.0
5 64.589411 22.223931 1 -1.0
6 43.758721 -20.830690 -1 -1.0
7 89.177300 45.418579 1 -1.0
8 96.366276 7.188976 1 -1.0
9 38.344152 -58.022124 -1 1.0
Result I Intend - Rolling window of 4(index 1,2,3,4) should be set and mode function be applied and result
should be assigned to index 0,then next rolling window(index 2,3,4,5) and result should
be assigned to index 1 and so on..

You have to reverse your list before then shift of 1 (because you don't want the current index in the result):
majority = lambda x: 0 if len((m := x.mode())) > 1 else m[0]
df['Voting'] = (df[::-1].rolling(4, min_periods=1)['Result']
.apply(majority).shift())
print(df)
# Output
Close Change Result Voting
0 54.881350 NaN 0 -1.0
1 71.518937 16.637586 1 -1.0
2 60.276338 -11.242599 -1 -1.0
3 54.488318 -5.788019 -1 0.0
4 42.365480 -12.122838 -1 1.0
5 64.589411 22.223931 1 0.0
6 43.758721 -20.830690 -1 1.0
7 89.177300 45.418579 1 0.0
8 96.366276 7.188976 1 -1.0
9 38.344152 58.022124 -1 NaN

Sample from dataframe with conditions

I have a large dataset and I want to sample from it but with a conditional. What I need is a new dataframe with the almost the same amount (count) of values of a boolean column of `0 and 1'
What I have:
df['target'].value_counts()
0 = 4000
1 = 120000
What I need:
new_df['target'].value_counts()
0 = 4000
1 = 6000
I know I can df.sample but I dont know how to insert the conditional.
Thanks

Since 1.1.0, you can use groupby.sample if you need the same number of rows for each group:
df.groupby('target').sample(4000)
Demo:
df = pd.DataFrame({'x': [0] * 10 + [1] * 25})
df.groupby('x').sample(5)
x
8 0
6 0
7 0
2 0
9 0
18 1
33 1
24 1
32 1
15 1
If you need to sample conditionally based on the group value, you can do:
df.groupby('target', group_keys=False).apply(
lambda g: g.sample(4000 if g.name == 0 else 6000)
)
Demo:
df.groupby('x', group_keys=False).apply(
lambda g: g.sample(4 if g.name == 0 else 6)
)
x
7 0
8 0
2 0
1 0
18 1
12 1
17 1
22 1
30 1
28 1

Assuming the following input and using the values 4/6 instead of 4000/6000:
df = pd.DataFrame({'target': [0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1]})
You could groupby your target and sample to take at most N values per group:
df.groupby('target', group_keys=False).apply(lambda g: g.sample(min(len(g), 6)))
example output:
target
4 0
0 0
8 0
12 0
10 1
14 1
1 1
7 1
11 1
13 1
If you want the same size you can simply use df.groupby('target').sample(n=4)

populate row with opposite value of the xth previous row, if condition is true

Following is the Dataframe I am starting from:
import pandas as pd
import numpy as np
d= {'PX_LAST':[1,2,3,3,3,1,2,1,1,1,3,3],'ma':[2,2,2,2,2,2,2,2,2,2,2,2],'action':[0,0,1,0,0,-1,0,1,0,0,-1,0]}
df_zinc = pd.DataFrame(data=d)
df_zinc
Now, I need to add a column called 'buy_sell', which:
when 'action'==1, populates with 1 if 'PX_LAST' >'ma', and with -1 if 'PX_LAST'<'ma'
when 'action'==-1, populates with the opposite of the previous non-zero value that was populated
FYI: in my data, the row that needs to be filled with the opposite of the previous non-zero item is always at the same distance from the previous non-zero item (i.e., 2 in the current example). This should facilitate making the code.
the code that I made so far is the following. It seems right to me. Do you have any fixes to propose?
while index < df_zinc.shape[0]:
if df_zinc['action'][index] == 1:
if df_zinc['PX_LAST'][index]<df_zinc['ma'][index]:
df_zinc.loc[index,'buy_sell'] = -1
else:
df_zinc.loc[index,'buy_sell'] = 1
elif df_zinc['action'][index] == -1:
df_zinc['buy_sell'][index] = df_zinc['buy_sell'][index-3]*-1
index=index+1
df_zinc
the resulting dataframe would look like this:
df_zinc['buy_sell'] = [0,0,1,0,0,-1,0,-1,0,0,1,0]
df_zinc

So, this would be my suggestion according to the example output (and assuming I understood the question properly:
def buy_sell(row):
if row['action'] == 0:
return 0
if row['PX_LAST'] > row['ma']:
return 1 * (-1 if row['action'] == 0 else 1)
else:
return -1 * (-1 if row['action'] == 0 else 1)
return 0
df_zinc = df_zinc.assign(buy_sell=df_zinc.apply(buy_sell, axis=1))
df_zinc
This should behave as expected by the rules. It does not take into account the possibility of 'PX_LAST' being equal to 'ma', returning 0 by default, as it was not clear what rule to follow in that scenario.
EDIT
Ok, after the new logic explained, I think this should do the trick:
def assign_buysell(df):
last_nonzero = None
def buy_sell(row):
nonlocal last_nonzero
if row['action'] == 0:
return 0
if row['action'] == 1:
if row['PX_LAST'] < row['ma']:
last_nonzero = -1
elif row['PX_LAST'] > row['ma']:
last_nonzero = 1
elif row['action'] == -1:
last_nonzero = last_nonzero * -1
return last_nonzero
return df.assign(buy_sell=df.apply(buy_sell, axis=1))
df_zinc = assign_buysell(df_zinc)
This solution is independent of how long ago the nonzero value was seen, it simply remembers the last nonzero value and pipes the opposite wen action is -1.

You can use np.select, and use np.nan as a label for the rows that satisfy the third condition:
c1 = df_zinc.action.eq(1) & df_zinc.PX_LAST.gt(df_zinc.ma)
c2 = df_zinc.action.eq(1) & df_zinc.PX_LAST.lt(df_zinc.ma)
c3 = df_zinc.action.eq(-1)
df_zinc['buy_sell'] = np.select([c1,c2, c3], [1, -1, np.nan])
Now in order to fill NaNs with the value from n rows above (in this case 3), you can fillna with a shifted version of the dataframe:
df_zinc['buy_sell'] = df_zinc.buy_sell.fillna(df_zinc.buy_sell.shift(3)*-1)
Output
PX_LAST ma action buy_sell
0 1 2 0 0.0
1 2 2 0 0.0
2 3 2 1 1.0
3 3 2 0 0.0
4 3 2 0 0.0
5 1 2 -1 -1.0
6 2 2 0 0.0
7 1 2 1 -1.0
8 1 2 0 0.0
9 1 2 0 0.0
10 3 2 -1 1.0
11 3 2 0 0.0

I would use np.select for this, since you have multiple conditions:
conditions = [
(df_zinc['action'] == 1) & (df_zinc['PX_LAST'] > df_zinc['ma']),
(df_zinc['action'] == 1) & (df_zinc['PX_LAST'] < df_zinc['ma']),
(df_zinc['action'] == -1) & (df_zinc['PX_LAST'] > df_zinc['ma']),
(df_zinc['action'] == -1) & (df_zinc['PX_LAST'] < df_zinc['ma'])
]
choices = [1, -1, 1, -1]
df_zinc['buy_sell'] = np.select(conditions, choices, default=0)
result
print(df_zinc)
PX_LAST ma action buy_sell
0 1 2 0 0
1 2 2 0 0
2 3 2 1 1
3 3 2 0 0
4 3 2 0 0
5 1 2 -1 -1
6 2 2 0 0
7 1 2 1 -1
8 1 2 0 0
9 1 2 0 0
10 3 2 -1 1
11 3 2 0 0

here my solution using the function shift() to trap the data of 3th up row:
df_zinc['buy_sell'] = 0
df_zinc.loc[(df_zinc['action'] == 1) & (df_zinc['PX_LAST'] < df_zinc['ma']), 'buy_sell'] = -1
df_zinc.loc[(df_zinc['action'] == 1) & (df_zinc['PX_LAST'] > df_zinc['ma']), 'buy_sell'] = 1
df_zinc.loc[df_zinc['action'] == -1, 'buy_sell'] = -df_zinc['buy_sell'].shift(3)
df_zinc['buy_sell'] = df_zinc['buy_sell'].astype(int)
print(df_zinc)
output:
PX_LAST ma action buy_sell
0 1 2 0 0
1 2 2 0 0
2 3 2 1 1
3 3 2 0 0
4 3 2 0 0
5 1 2 -1 -1
6 2 2 0 0
7 1 2 1 -1
8 1 2 0 0
9 1 2 0 0
10 3 2 -1 1
11 3 2 0 0

substitute all numbers in a matrix with equivalent letters when Na exists

There is a huge matrix whose elements are numbers in the range of 1 to 15. I want to transform the matrix to the one whose elements be letters such that 1 becomes "a", 2 becomes "b", and so on. Finally I want to merge each row and create a sequence of it. As a simple example:
import pandas as pd
import numpy as np, numpy.random
numpy.random.seed(1)
A = pd.DataFrame (np.random.randint(1,16,10).reshape(2,5))
A.iloc[1,4]= np.NAN
A
# 0 1 2 3 4
#0 6 12 13 9 10.0
#1 12 6 1 1 NaN
If there were no Na in the dataset, I would use this code:
pd.DataFrame(list(map(''.join, A.applymap(lambda n: chr(n + 96)).as_matrix())))
Here, it gives this error:
TypeError: ('integer argument expected, got float', 'occurred at index 4')
The expected output is:
0
0 flmij
1 lfaa
The first row should have 5 elements and the second one should have 4 elements.

Use if-else condition with sum:
df = pd.DataFrame(A.applymap(lambda n: chr(int(n) + 96) if pd.notnull(n) else '')
.values.sum(axis=1))
print (df)
0
0 flmij
1 lfaa
Details:
print (A.applymap(lambda n: chr(int(n) + 96) if pd.notnull(n) else ''))
0 1 2 3 4
0 f l m i j
1 l f a a
print (A.applymap(lambda n: chr(int(n) + 96) if pd.notnull(n) else '').values)
[['f' 'l' 'm' 'i' 'j']
['l' 'f' 'a' 'a' '']]
print (A.applymap(lambda n: chr(int(n) + 96) if pd.notnull(n) else '').values.sum(axis=1))
['flmij' 'lfaa']
Another solution:
print (A.stack().astype(int).add(96).apply(chr).sum(level=0))
0 flmij
1 lfaa
dtype: object
Details:
Reshape to Series:
print (A.stack())
0 0 6.0
1 12.0
2 13.0
3 9.0
4 10.0
1 0 12.0
1 6.0
2 1.0
3 1.0
dtype: float64
Convert to integers:
print (A.stack().astype(int))
0 0 6
1 12
2 13
3 9
4 10
1 0 12
1 6
2 1
3 1
dtype: int32
Add number:
print (A.stack().astype(int).add(96))
0 0 102
1 108
2 109
3 105
4 106
1 0 108
1 102
2 97
3 97
dtype: int32
Convert to letters:
print (A.stack().astype(int).add(96).apply(chr))
0 0 f
1 l
2 m
3 i
4 j
1 0 l
1 f
2 a
3 a
dtype: object
Sum by first level of MultiIndex:
print (A.stack().astype(int).add(96).apply(chr).sum(level=0))
0 flmij
1 lfaa
dtype: object

try this,
A.fillna(0,inplace=True)
A.applymap(lambda x: (chr(int(x) + 96))).sum(axis=1).str.replace('`','')
0 flmij
1 lfaa
dtype: object

Could use a categorical. Useful if you're doing more than just mapping to individual characters.
import pandas as pd
import numpy as np, numpy.random
numpy.random.seed(1)
A_int = pd.DataFrame(np.random.randint(1,16,10).reshape(2,5))
A_int.iloc[1,4]= np.NAN
int_vals = list(range(1,16))
chr_vals = [chr(n+96) for n in int_vals]
A_chr = A_int.apply(axis=0, func=lambda x: pd.Categorical(x, categories=int_vals, ordered=True).rename_categories(chr_vals))
A_chr.apply(axis=1, func=lambda x: ''.join([str(i) for i in x[pd.notnull(x)]]))

Groupby on condition and calculate sum of subgroups

Here is my data:
import numpy as np
import pandas as pd
z = pd.DataFrame({'a':[1,1,1,2,2,3,3],'b':[3,4,5,6,7,8,9], 'c':[10,11,12,13,14,15,16]})
z
a b c
0 1 3 10
1 1 4 11
2 1 5 12
3 2 6 13
4 2 7 14
5 3 8 15
6 3 9 16
Question:
How can I do calculation on different element of each subgroup? For example, for each group, I want to extract any element in column 'c' which its corresponding element in column 'b' is between 4 and 9, and sum them all.
Here is the code I wrote: (It runs but I cannot get the correct result)
gbz = z.groupby('a')
# For displaying the groups:
gbz.apply(lambda x: print(x))
list = []
def f(x):
list_new = []
for row in range(0,len(x)):
if (x.iloc[row,0] > 4 and x.iloc[row,0] < 9):
list_new.append(x.iloc[row,1])
list.append(sum(list_new))
results = gbz.apply(f)
The output result should be something like this:
a c
0 1 12
1 2 27
2 3 15

It might just be easiest to change the order of operations, and filter against your criteria first - it does not change after the groupby.
z.query('4 < b < 9').groupby('a', as_index=False).c.sum()
which yields
a c
0 1 12
1 2 27
2 3 15

Use
In [2379]: z[z.b.between(4, 9, inclusive=False)].groupby('a', as_index=False).c.sum()
Out[2379]:
a c
0 1 12
1 2 27
2 3 15
Or
In [2384]: z[(4 < z.b) & (z.b < 9)].groupby('a', as_index=False).c.sum()
Out[2384]:
a c
0 1 12
1 2 27
2 3 15

You could also groupby first.
z = z.groupby('a').apply(lambda x: x.loc[x['b']\
.between(4, 9, inclusive=False), 'c'].sum()).reset_index(name='c')
z
a c
0 1 12
1 2 27
2 3 15

Or you can use
z.groupby('a').apply(lambda x : sum(x.loc[(x['b']>4)&(x['b']<9),'c']))\
.reset_index(name='c')
Out[775]:
a c
0 1 12
1 2 27
2 3 15