Develop Reference

Flag creation based on count of consecutive ones in a column

I have a data frame with a column with only 0's and 1's. I need to create a flag column where there are more than a certain number of consecutive ones in the first column.
In the example below, x >= 4 , if there are 4 or more consecutive one's, then the flag should be 1 for all those consecutive rows.
col1 Flag
0 1 0
1 0 0
2 1 1
3 1 1
4 1 1
5 1 1
6 0 0
7 1 0
8 1 0
9 0 0
10 1 1
11 1 1
12 1 1
13 1 1
14 1 1
15 0 0
One change, let's say there is a new column group, we need to group by that and find the flag,
Group col1 Flag
0 A 1 0
1 B 0 0
2 B 1 1
3 B 1 1
4 B 1 1
5 B 1 1
6 C 0 0
7 C 1 0
8 C 1 0
9 C 0 0
10 D 1 0
11 D 1 0
12 D 1 0
13 E 1 0
14 E 1 0
15 E 0 0
As you can there are consecutive ones from 10 to 14 but they belong to different groups. And elements in group can be in any order.

No that hard try with cumsum create the key then do the transform count
(df.groupby(df.col1.ne(1).cumsum())['col1'].transform('count').ge(5) & df.col1.eq(1)).astype(int)
Out[83]:
0 0
1 0
2 1
3 1
4 1
5 1
6 0
7 0
8 0
9 0
10 1
11 1
12 1
13 1
14 1
15 0
Name: col1, dtype: int32

You can achieve this in a couple of steps:
rolling(4).sum() to attain consecutive summations of your column
Use where to get the 1's from "col1" where their summation window (from the previous step) is >= 4. Turn the rest of the values into np.NaN
bfill(limit=3) to backwards fill the leftover 1s in your column by a maximum of 3 places.
fillna(0) fill what's leftover with 0
df["my_flag"] = (df["col1"]
.where(
df["col1"].rolling(4).sum() >= 4
) # Selects the 1's whose consecutive sum >= 4. All other values become NaN
.bfill(limit=3) # Moving backwards from our leftover values,
# take the existing value and fill in a maximum of 3 NaNs
.fillna(0) # Fill in the rest of the NaNs with 0
.astype(int)) # Cast to integer data type, since we were working with floats temporarily
print(df)
col1 Flag my_flag
0 1 0 0
1 0 0 0
2 1 1 1
3 1 1 1
4 1 1 1
5 1 1 1
6 0 0 0
7 1 0 0
8 1 0 0
9 0 0 0
10 1 1 1
11 1 1 1
12 1 1 1
13 1 1 1
14 1 1 1
15 0 0 0
Edit:
For a grouped approach, you just need to use groupby().rolling to create your mask for use in where(). Everything after that is the same. I separated the rolling step to keep it as readable as possible:
grouped_counts_ge_4 = (df.groupby("Group")["col1"]
.rolling(4)
.sum()
.ge(4)
.reset_index(level=0, drop=True))
df["my_flag"] = (df["col1"]
.where(grouped_counts_ge_4)
.bfill(limit=3) # Moving backwards from our leftover values, take the existing value and fill in a maximum of 3 NaNs
.fillna(0) # Fill in the rest of the NaNs with 0
.astype(int)) # Cast to integer data type, since we were working with floats temporarily
print(df)
Group col1 Flag my_flag
0 A 1 0 0
1 B 0 0 0
2 B 1 1 1
3 B 1 1 1
4 B 1 1 1
5 B 1 1 1
6 C 0 0 0
7 C 1 0 0
8 C 1 0 0
9 C 0 0 0
10 D 1 0 0
11 D 1 0 0
12 D 1 0 0
13 E 1 0 0
14 E 1 0 0
15 E 0 0 0

Try this:
df['Flag'] = np.where(df['col1'].groupby((df['col1'].diff().ne(0) | df['col1'].eq(0)).cumsum()).transform('size').ge(4),1,0)

How to concatenate bit columns in Python Pandas?

Seems like an easy question but I'm running into an odd error. I have a large dataframe with 24+ columns that all contain 1s or 0s. I wish to concatenate each field to create a binary key that'll act as a signature.
However, when the number of columns exceeds 12, the whole process falls apart.
a = np.zeros(shape=(3,12))
df = pd.DataFrame(a)
df = df.astype(int) # This converts each 0.0 into just 0
df[2]=1 # Changes one column to all 1s
#result
0 1 2 3 4 5 6 7 8 9 10 11
0 0 0 1 0 0 0 0 0 0 0 0 0
1 0 0 1 0 0 0 0 0 0 0 0 0
2 0 0 1 0 0 0 0 0 0 0 0 0
Concatenating function...
df['new'] = df.astype(str).sum(1).astype(int).astype(str) # Concatenate
df['new'].apply('{0:0>12}'.format) # Pad leading zeros
# result
0 1 2 3 4 5 6 7 8 9 10 11 new
0 0 0 1 0 0 0 0 0 0 0 0 0 001000000000
1 0 0 1 0 0 0 0 0 0 0 0 0 001000000000
2 0 0 1 0 0 0 0 0 0 0 0 0 001000000000
This is good. However, if I increase the number of columns to 13, I get...
a = np.zeros(shape=(3,13))
# ...same intermediate steps as above...
0 1 2 3 4 5 6 7 8 9 10 11 12 new
0 0 0 1 0 0 0 0 0 0 0 0 0 0 00-2147483648
1 0 0 1 0 0 0 0 0 0 0 0 0 0 00-2147483648
2 0 0 1 0 0 0 0 0 0 0 0 0 0 00-2147483648
Why am I getting -2147483648? I was expecting 0010000000000
Any help is appreciated!

Convert the last non-zero value to 0 for each row in a pandas DataFrame

I'm trying to modify my data frame in a way that the last variable of a label encoded feature is converted to 0. For example, I have this data frame, top row being the labels and the first column as the index:
df
1 2 3 4 5 6 7 8 9 10
0 0 1 0 0 0 0 0 0 1 1
1 0 0 0 1 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 1 0
Columns 1-10 are the ones that have been encoded. What I want to convert this data frame to, without changing anything else is:
1 2 3 4 5 6 7 8 9 10
0 0 1 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
So the last values occurring in each row should be converted to 0. I was thinking of using the last_valid_index method, but that would take in the other remaining columns and change that as well, which I don't want. Any help is appreciated

You can use cumsum to build a boolean mask, and set to zero.
v = df.cumsum(axis=1)
df[v.lt(v.max(axis=1), axis=0)].fillna(0, downcast='infer')
1 2 3 4 5 6 7 8 9 10
0 0 1 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
Another similar option is reversing before calling cumsum, you can now do this in a single line.
df[~df.iloc[:, ::-1].cumsum(1).le(1)].fillna(0, downcast='infer')
1 2 3 4 5 6 7 8 9 10
0 0 1 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
If you have more columns, just apply these operations on the slice. Later, assign back.
u = df.iloc[:, :10]
df[u.columns] = u[~u.iloc[:, ::-1].cumsum(1).le(1)].fillna(0, downcast='infer')

Create Pandas DataFrame from (row, column, value) data

I have a Pandas Dataframe with three columns: row, column, value. The row values are all integers below some N, and the column values are all integers below some M. The values are all positive integers.
How do I efficiently create a Dataframe with N rows and M columns, with at index i, j the value val if (i, j , val) is a row in my original Dataframe, and some default value (0) otherwise? Furthermore, is it possible to create a sparse Dataframe immediately, since the data is already quite large, but N*M is still about 10 times the size of my data?

A NumPy solution would suit here for performance -
a = df.values
m,n = a[:,:2].max(0)+1
out = np.zeros((m,n),dtype=a.dtype)
out[a[:,0], a[:,1]] = a[:,2]
df_out = pd.DataFrame(out)
Sample run -
In [58]: df
Out[58]:
row col val
0 7 1 30
1 3 3 0
2 4 8 30
3 5 8 18
4 1 3 6
5 1 6 48
6 0 2 6
7 4 7 6
8 5 0 48
9 8 1 48
10 3 2 12
11 6 8 18
In [59]: df_out
Out[59]:
0 1 2 3 4 5 6 7 8
0 0 0 6 0 0 0 0 0 0
1 0 0 0 6 0 0 48 0 0
2 0 0 0 0 0 0 0 0 0
3 0 0 12 0 0 0 0 0 0
4 0 0 0 0 0 0 0 6 30
5 48 0 0 0 0 0 0 0 18
6 0 0 0 0 0 0 0 0 18
7 0 30 0 0 0 0 0 0 0
8 0 48 0 0 0 0 0 0 0

Finding efficiently pandas (part of) rows with unique values

Given a pandas dataframe with a row per individual/record. A row includes a property value and its evolution across time (0 to N).
A schedule includes the estimated values of a variable 'property' for a number of entities from day 1 to day 10 in the following example.
I want to filter entities with unique values for a given period and get those values
csv=',property,1,2,3,4,5,6,7,8,9,10\n0,100011,0,0,0,0,3,3,3,3,3,0\n1,100012,0,0,0,0,2,2,2,8,8,0\n2, \
100012,0,0,0,0,2,2,2,2,2,0\n3,100012,0,0,0,0,0,0,0,0,0,0\n4,100011,0,0,0,0,2,2,2,2,2,0\n5, \
180011,0,0,0,0,2,2,2,2,2,0\n6,110012,0,0,0,0,0,0,0,0,0,0\n7,110011,0,0,0,0,3,3,3,3,3,0\n8, \
110012,0,0,0,0,3,3,3,3,3,0\n9,110013,0,0,0,0,0,0,0,0,0,0\n10,100011,0,0,0,0,3,3,3,3,4,0'
from StringIO import StringIO
import numpy as np
schedule = pd.read_csv(StringIO(csv), index_col=0)
print schedule
property 1 2 3 4 5 6 7 8 9 10
0 100011 0 0 0 0 3 3 3 3 3 0
1 100012 0 0 0 0 2 2 2 8 8 0
2 100012 0 0 0 0 2 2 2 2 2 0
3 100012 0 0 0 0 0 0 0 0 0 0
4 100011 0 0 0 0 2 2 2 2 2 0
5 180011 0 0 0 0 2 2 2 2 2 0
6 110012 0 0 0 0 0 0 0 0 0 0
7 110011 0 0 0 0 3 3 3 3 3 0
8 110012 0 0 0 0 3 3 3 3 3 0
9 110013 0 0 0 0 0 0 0 0 0 0
10 100011 0 0 0 0 3 3 3 3 4 0
I want to find records/individuals for who property has not changed during a given period and the corresponding unique values
Here is what i came with : I want to locate individuals with property in [100011, 100012, 1100012] between days 7 and 10
props = [100011, 100012, 1100012]
begin = 7
end = 10
res = schedule['property'].isin(props)
df = schedule.ix[res, begin:end]
print "df \n%s " %df
We have :
df
7 8 9
0 3 3 3
1 2 8 8
2 2 2 2
3 0 0 0
4 2 2 2
10 3 3 4
res = df.apply(lambda x: np.unique(x).size == 1, axis=1)
print "res : %s\n" %res
df_f = df.ix[res,]
print "df filtered %s \n" % df_f
res = pd.Series(df_f.values.ravel()).unique().tolist()
print "unique values : %s " %res
Giving :
res :
0 True
1 False
2 True
3 True
4 True
10 False
dtype: bool
df filtered
7 8 9
0 3 3 3
2 2 2 2
3 0 0 0
4 2 2 2
unique values : [3, 2, 0]
As those operations need to be run many times (in millions) on a million rows dataframe, i need to be able to run it as quickly as possible.
(#MaxU) : schedule can be seen as a database/repository updated many times. The repository is then requested as well many times for unique values
Would you have some ideas for improvements/ alternate ways ?

Given your df
7 8 9
0 3 3 3
1 2 8 8
2 2 2 2
3 0 0 0
4 2 2 2
10 3 3 4
You can simplify your code to:
df_f = df[df.apply(pd.Series.nunique, axis=1) == 1]
print(df_f)
7 8 9
0 3 3 3
2 2 2 2
3 0 0 0
4 2 2 2
And the final step to:
res = df_f.iloc[:,0].unique().tolist()
print(res)
[3, 2, 0]
It's not fully vectorised, but maybe this clarifies things a bit towards that?