I want to create a new column which is a result of a shift function applied to a grouped values.
df = pd.DataFrame({'X': [0,1,0,1,0,1,0,1], 'Y':[2,4,3,1,2,3,4,5]})
df
X Y
0 0 2
1 1 4
2 0 3
3 1 1
4 0 2
5 1 3
6 0 4
7 1 5
def func(x):
x['Z'] = test['Y']-test['Y'].shift(1)
return x
df_new = df.groupby('X').apply(func)
X Y Z
0 0 2 NaN
1 1 4 2.0
2 0 3 -1.0
3 1 1 -2.0
4 0 2 1.0
5 1 3 1.0
6 0 4 1.0
7 1 5 1.0
As you can see from the output the values are shifted sequentally without accounting for a group by.
I have seen a similar question, but I could not figure out why it does not work as expected.
Python Pandas: how to add a totally new column to a data frame inside of a groupby/transform operation
The values are shifted without accounting for the groups because your func uses test (presumably some other object, likely another name for what you call df) directly instead of simply the group x.
def func(x):
x['Z'] = x['Y']-x['Y'].shift(1)
return x
gives me
In [8]: df_new
Out[8]:
X Y Z
0 0 2 NaN
1 1 4 NaN
2 0 3 1.0
3 1 1 -3.0
4 0 2 -1.0
5 1 3 2.0
6 0 4 2.0
7 1 5 2.0
but note that in this particular case you don't need to write a custom function, you can just call diff on the groupby object directly. (Of course other functions you might want to work with may be more complicated).
In [13]: df_new["Z2"] = df.groupby("X")["Y"].diff()
In [14]: df_new
Out[14]:
X Y Z Z2
0 0 2 NaN NaN
1 1 4 NaN NaN
2 0 3 1.0 1.0
3 1 1 -3.0 -3.0
4 0 2 -1.0 -1.0
5 1 3 2.0 2.0
6 0 4 2.0 2.0
7 1 5 2.0 2.0
Related
I have the data as below, the new pandas version doesn't preserve the grouped columns after the operation of fillna/ffill/bfill. Is there a way to have the grouped data?
data = """one;two;three
1;1;10
1;1;nan
1;1;nan
1;2;nan
1;2;20
1;2;nan
1;3;nan
1;3;nan"""
df = pd.read_csv(io.StringIO(data), sep=";")
print(df)
one two three
0 1 1 10.0
1 1 1 NaN
2 1 1 NaN
3 1 2 NaN
4 1 2 20.0
5 1 2 NaN
6 1 3 NaN
7 1 3 NaN
print(df.groupby(['one','two']).ffill())
three
0 10.0
1 10.0
2 10.0
3 NaN
4 20.0
5 20.0
6 NaN
7 NaN
With the most recent pandas if we would like keep the groupby columns , we need to adding apply here
out = df.groupby(['one','two']).apply(lambda x : x.ffill())
Out[219]:
one two three
0 1 1 10.0
1 1 1 10.0
2 1 1 10.0
3 1 2 NaN
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
Does it what you expect?
df['three']= df.groupby(['one','two'])['three'].ffill()
print(df)
# Output:
one two three
0 1 1 10.0
1 1 1 10.0
2 1 1 10.0
3 1 2 NaN
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
Yes please set the index and then try grouping it so that it will preserve the columns as shown here:
df = pd.read_csv(io.StringIO(data), sep=";")
df.set_index(['one','two'], inplace=True)
df.groupby(['one','two']).ffill()
I have a column. how I can make a new column to count repetative positive and negative signs?
col1
-5
-3
-7
4
5
-0.5
6
8
9
col1 count_sign
-5 3
-3 3
-7 3
4 2
5 2
-0.5 1
6 3
8 3
9 3
the first 3 rows are 3 because we have 3 negative signs in the first 3 rows, then 2 positive signs and ....
# identify the change of signs among rows,
# making count as NaN, where sign is same, else 1
df['count']=np.where(np.sign(df['col1']).diff().eq(0),
np.nan,
1)
# cumsum to group the rows
df['count']=df['count'].cumsum().ffill()
# groupby to take count of each group of rows and return groupsize using transform
df['count']=df.groupby('count')['col1'].transform('size')
df
col1 count
0 -5.0 3
1 -3.0 3
2 -7.0 3
3 4.0 2
4 5.0 2
5 -0.5 1
6 6.0 3
7 8.0 3
8 9.0 3
To add a sign to the count values
df['count']=np.where(np.sign(df['col1']).diff().eq(0),
np.nan,
1)
df['count']=df['count'].cumsum().ffill()
df['count']=df.groupby('count')['col1'].transform('size')*np.sign(df['col1'])
df
col1 count
0 -5.0 -3.0
1 -3.0 -3.0
2 -7.0 -3.0
3 4.0 2.0
4 5.0 2.0
5 -0.5 -1.0
6 6.0 3.0
7 8.0 3.0
8 9.0 3.0
I need to compute lagged means per groups in my dataframe. This is how my df looks like:
name value round
0 a 5 3
1 b 4 3
2 c 3 2
3 d 1 2
4 a 2 1
5 c 1 1
0 c 1 3
1 d 4 3
2 b 3 2
3 a 1 2
4 b 5 1
5 d 2 1
I would like to compute lagged means for column value per name and round. That is, for name a in round 3 I need to have value_mean = 1.5 (because (1+2)/2). And of course, there will be nan values when round = 1.
I tried this:
df['value_mean'] = df.groupby('name').expanding().mean().groupby('name').shift(1)['value'].values
but it gives a nonsense:
name value round value_mean
0 a 5 3 NaN
1 b 4 3 5.0
2 c 3 2 3.5
3 d 1 2 NaN
4 a 2 1 4.0
5 c 1 1 3.5
0 c 1 3 NaN
1 d 4 3 3.0
2 b 3 2 2.0
3 a 1 2 NaN
4 b 5 1 1.0
5 d 2 1 2.5
Any idea, how can I do this, please? I found this, but it seems not relevant for my problem: Calculate the mean value using two columns in pandas
You can do that as follows
# sort the values as they need to be counted
df.sort_values(['name', 'round'], inplace=True)
df.reset_index(drop=True, inplace=True)
# create a grouper to calculate the running count
# and running sum as the basis of the average
grouper= df.groupby('name')
ser_sum= grouper['value'].cumsum()
ser_count= grouper['value'].cumcount()+1
ser_mean= ser_sum.div(ser_count)
ser_same_name= df['name'] == df['name'].shift(1)
# finally you just have to set the first entry
# in each name-group to NaN (this usually would
# set the entries for each name and round=1 to NaN)
df['value_mean']= ser_mean.shift(1).where(ser_same_name, np.NaN)
# if you want to see the intermediate products,
# you can uncomment the following lines
#df['sum']= ser_sum
#df['count']= ser_count
df
Output:
name value round value_mean
0 a 2 1 NaN
1 a 1 2 2.0
2 a 5 3 1.5
3 b 5 1 NaN
4 b 3 2 5.0
5 b 4 3 4.0
6 c 1 1 NaN
7 c 3 2 1.0
8 c 1 3 2.0
9 d 2 1 NaN
10 d 1 2 2.0
11 d 4 3 1.5
I have a panda dataframe like this:
second block
0 1 a
1 2 b
2 3 c
3 4 a
4 5 c
This is a sequential data and I would like to get a new column which is the time difference between the current block and next time it repeats.
second block freq
0 1 a 3 //(4-1)
1 2 b 0 //(not repeating)
2 3 c 2 //(5-3)
3 4 a 0 //(not repeating)
4 5 c 0 //(not repeating)
I have tried to get the unique list of blocks. Then a for loop that do as below.
for i in unique_block:
df['freq'] = df['timestamp'].shift(-1) - df['timestamp']
I do not know how to get 0 for row index 1,3,4 and since the dataframe is too big. This is not efficient. This is not working.
Thanks.
Use groupby + diff(periods=-1). Multiply by -1 to get your difference convention and fillna with 0.
df['freq'] = (df.groupby('block').diff(-1)*-1).fillna(0)
second block freq
0 1 a 3.0
1 2 b 0.0
2 3 c 2.0
3 4 a 0.0
4 5 c 0.0
You can use shift and transform in your groupby:
df['freq'] = df.groupby('block').second.transform(lambda x: x.shift(-1) - x).fillna(0)
>>> df
second block freq
0 1 a 3.0
1 2 b 0.0
2 3 c 2.0
3 4 a 0.0
4 5 c 0.0
Using
df.groupby('block').second.apply(lambda x : x.diff().shift(-1)).fillna(0)
Out[242]:
0 3.0
1 0
2 2.0
3 0
4 0
Name: second, dtype: float64
I have one table with 1000s of rows that looks like this:
file1:
apples1 + hate 0 0 0 2 4 6 0 1
apples2 + hate 0 2 0 4 4 6 0 2
apples4 + hate 0 2 0 4 4 6 0 2
and another file with same headers in file2 - nb some headers are missing in file1:
apples1 + hate 0 0 0 1 4 6 0 2
apples2 + hate 0 1 0 6 4 6 0 2
apples3 + hate 0 2 0 4 4 6 0 2
apples4 + hate 0 1 0 3 4 3 0 1
I want to compare the two files in pandas and average across common columns. I do not want to print columns that are in one file only. So the resultant file would look like:
apples1 + hate 0 0 0 1.5 4 6 0 1.5
apples2 + hate 0 1.5 0 5 4 6 0 2
apples4 + hate 0 2 0 3.5 4 6 0 2
There are two steps in this solution.
concatenate all your dataframes by stacking them vertically (axis=0, the default) using pandas.concat(...) and specifying a join of 'inner' to only maintain columns that in all the dataframes.
call mean(...) function on resultant dataframe.
Example:
In [1]: df1 = pd.DataFrame([[1,2,3], [4,5,6]], columns=['a','b','c'])
In [2]: df2 = pd.DataFrame([[1,2],[3,4]], columns=['a','c'])
In [3]: df1
Out[3]:
a b c
0 1 2 3
1 4 5 6
In [4]: df2
Out[4]:
a c
0 1 2
1 3 4
In [5]: df3 = pd.concat([df1, df2], join='inner')
In [6]: df3
Out[6]:
a c
0 1 3
1 4 6
0 1 2
1 3 4
In [7]: df3.mean()
Out[7]:
a 2.25
c 3.75
dtype: float64
Let's try this:
df1 = pd.read_csv('file1', header=None)
df2 = pd.read_csv('file2', header=None)
Set index to first three columns ie.. "apple1 + hate"
df1 = df1.set_index([0,1,2])
df2 = df2.set_index([0,1,2])
Let's use merge to inner join datafiles on indexes, and the groupby columns with the same name and aggregate with mean:
df1.merge(df2, right_index=True, left_index=True)\
.pipe(lambda x: x.groupby(x.columns.str.extract('(\w+)\_[xy]', expand=False),
axis=1, sort=False).mean()).reset_index()
Output:
0 1 2 3 4 5 6 7 8 9 10
0 apples1 + hate 0.0 0.0 0.0 1.5 4.0 6.0 0.0 1.5
1 apples2 + hate 0.0 1.5 0.0 5.0 4.0 6.0 0.0 2.0
2 apples4 + hate 0.0 1.5 0.0 3.5 4.0 4.5 0.0 1.5