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()
Related
I have the following dataframe, observations are grouped in pairs. NaN here represents different products traded in pair wrt A. I want to groupby transaction and compute
A/NaN so that the value for all NaNs can be expressed in unit A.
transaction name value ...many other columns
1 A 3
1 NaN 5
2 NaN 7
2 A 6
3 A 4
3 NaN 3
4 A 10
4 NaN 9
5 C 8
5 A 6
..
Thus the desired df would be
transaction name value new_column ...many other columns
1 A 3 NaN
1 NaN 6 0.5
2 NaN 7 0.8571
2 A 6 NaN
3 A 4 1.333
3 NaN 3 NaN
4 A 10 1.111
4 NaN 9 NaN
5 C 8 0.75
5 A 6 NaN
...
First filter rows with A and convert transaction to index for possible divide rows with missing value by mapped transaction by Series.map:
m = df['name'].ne('A')
s = df[~m].set_index('transaction')['value']
df.loc[m, 'new_column'] = df.loc[m, 'transaction'].map(s) / df.loc[m, 'value']
print (df)
transaction name value new_column
0 1 A 3 NaN
1 1 NaN 5 0.600000
2 2 NaN 7 0.857143
3 2 A 6 NaN
4 3 A 4 NaN
5 3 NaN 3 1.333333
6 4 A 10 NaN
7 4 NaN 9 1.111111
8 5 NaN 8 0.750000
9 5 A 6 NaN
EDIT: There is multiple A values per groups, not only one, possible solution is removed duplicates:
print (df)
transaction name value
0 1 A 3
1 1 A 4
2 1 NaN 5
3 2 NaN 7
4 2 A 6
5 3 A 4
6 3 NaN 3
7 4 A 10
8 4 NaN 9
9 5 C 8
10 5 A 6
# s = df[~m].set_index('transaction')['value']
# df.loc[m, 'new_column'] = df.loc[m, 'transaction'].map(s) / df.loc[m, 'value']
# print (df)
#InvalidIndexError: Reindexing only valid with uniquely valued Index objects
m = df['name'].ne('A')
print (df[~m].drop_duplicates(['transaction','name']))
transaction name value
0 1 A 3
4 2 A 6
5 3 A 4
7 4 A 10
10 5 A 6
s = df[~m].drop_duplicates(['transaction','name']).set_index('transaction')['value']
df.loc[m, 'new_column'] = df.loc[m, 'transaction'].map(s) / df.loc[m, 'value']
print (df)
transaction name value new_column
0 1 A 3 NaN <- 2 times a per 1 group
1 1 A 4 NaN <- 2 times a per 1 group
2 1 NaN 5 0.600000
3 2 NaN 7 0.857143
4 2 A 6 NaN
5 3 A 4 NaN
6 3 NaN 3 1.333333
7 4 A 10 NaN
8 4 NaN 9 1.111111
9 5 C 8 0.750000
10 5 A 6 NaN
Assuming there are only two values per transaction, you can use agg and divide the first and last element by each other:
df.loc[df['name'].isna(), 'new_column'] = df.sort_values(by='name').\
groupby('transaction')['value'].\
agg(f='first', l='last').agg(lambda x: x['f'] / x['l'], axis=1)
I have a following problem.
I want to compute mean of last 2 observations per name and round and lag it. See following example:
df = pd.DataFrame(data={ 'name':["a","a","a","a","b","b","c" ] , 'value':[6,5,4,3,1,2,1] ,
'round':[1,2,3,4,1,2,1 ]})
Desired output is:
df = pd.DataFrame(data={ 'name':["a","a","a","a","b","b","c" ] , 'value':[6,5,4,3,1,2,1] ,
'round':[1,2,3,4,1,2,1 ], 'mean_last_2':["NaN","NaN",5.5,4.5,"NaN","NaN","NaN"]})
I tried this, but got "AttributeError: 'float' object has no attribute 'shift'":
df['mean_last_2'] = df.groupby("name")['value'].apply(lambda x:
x.tail(2).mean().shift(1))
How can I fix it please?
You could try something like this:
df['mean_last_2'] = df.groupby('name')['value'].apply(lambda x: x.rolling(2).mean().shift())
Output:
name value round mean_last_2
0 a 6 1 NaN
1 a 5 2 NaN
2 a 4 3 5.5
3 a 3 4 4.5
4 b 1 1 NaN
5 b 2 2 NaN
6 c 1 1 NaN
You can do something like
df.groupby("name").apply(lambda d: d.assign(mean_last_2 = d['value'].rolling(2).mean().shift()))
to get
name value round mean_last_2
name
a 0 a 6 1 NaN
1 a 5 2 NaN
2 a 4 3 5.5
3 a 3 4 4.5
b 4 b 1 1 NaN
5 b 2 2 NaN
c 6 c 1 1 NaN
I currently have a dataframe which looks like this:
df:
store item sales
0 1 1 10
1 1 2 20
2 2 1 10
3 3 2 20
4 4 3 10
5 3 4 15
...
I wanted to view the total sales of each items for each store so I used pivot table to create this:
p_table = pd.pivot_table(df, index='store', values='sales', columns='item', aggfunc=np.sum)
which gives something like:
sales
item 1 2 3 4
store
1 20 30 10 8
2 10 14 12 13
3 1 23 29 10
....
What I want to do now is apply some functions so that each total sales of items represents the percentage of the total sales for a particular store. For example, the value for item 1 at store1 would become:
1. 20/(20+30+10+8) * 100
I am struggling to do this for stacked dataframe. Any suggestions would be much appreciated.
Thanks
I think need divide by div with Series created by sum:
print (p_table)
item 1 2 3 4
store
1 10.0 20.0 NaN NaN
2 10.0 NaN NaN NaN
3 NaN 20.0 NaN 15.0
4 NaN NaN 10.0 NaN
print (p_table.sum(axis=1))
store
1 30.0
2 10.0
3 35.0
4 10.0
dtype: float64
out = p_table.div(p_table.sum(axis=1), axis=0)
print (out)
item 1 2 3 4
store
1 0.333333 0.666667 NaN NaN
2 1.000000 NaN NaN NaN
3 NaN 0.571429 NaN 0.428571
4 NaN NaN 1.0 NaN
I have dataframe look like this:
raw_data ={'col0':[1,4,5,1,3,3,1,5,8,9,1,2]}
df = DataFrame(raw_data)
col0
0 1
1 4
2 5
3 1
4 3
5 3
6 1
7 5
8 8
9 9
10 1
11 2
What I want to do is to count every 3 rows to fit condition(df['col0']>3) and make new col looks like this:
col0 col_roll_count3
0 1 0
1 4 1
2 5 2 #[index 0,1,2/ 4,5 fit the condition]
3 1 2
4 3 1
5 3 0 #[index 3,4,5/no fit the condition]
6 1 0
7 5 1
8 8 2
9 9 3
10 1 2
11 2 1
How can I achieve that?
I tried this but failed:
df['col_roll_count3'] = df[df['col0']>3].rolling(3).count()
print(df)
col0 col1
0 1 NaN
1 4 1.0
2 5 2.0
3 1 NaN
4 3 NaN
5 3 NaN
6 1 NaN
7 5 3.0
8 8 3.0
9 9 3.0
10 1 NaN
11 2 NaN
df['col_roll_count3'] = df['col0'].gt(3).rolling(3).sum()
Let's use rolling, apply, np.count_nonzero:
df['col_roll_count3'] = df.col0.rolling(3,min_periods=1)\
.apply(lambda x: np.count_nonzero(x>3))
Output:
col0 col_roll_count3
0 1 0.0
1 4 1.0
2 5 2.0
3 1 2.0
4 3 1.0
5 3 0.0
6 1 0.0
7 5 1.0
8 8 2.0
9 9 3.0
10 1 2.0
11 2 1.0
I managed to solve using if and for loops but I'm looking for a less computationally expensive way to do this. i.e. using apply or map or any other technique
d = {1:10, 2:20, 3:30}
df
a b
1 35
1 nan
1 nan
2 nan
2 47
2 nan
3 56
3 nan
I want to fill missing values of column b according to dict d, i.e. output should be
a b
1 35
1 10
1 10
2 20
2 47
2 20
3 56
3 30
You can use fillna or combine_first by maped a column:
print (df['a'].map(d))
0 10
1 10
2 10
3 20
4 20
5 20
6 30
7 30
Name: a, dtype: int64
df['b'] = df['b'].fillna(df['a'].map(d))
print (df)
a b
0 1 35.0
1 1 10.0
2 1 10.0
3 2 20.0
4 2 47.0
5 2 20.0
6 3 56.0
7 3 30.0
df['b'] = df['b'].combine_first(df['a'].map(d))
print (df)
a b
0 1 35.0
1 1 10.0
2 1 10.0
3 2 20.0
4 2 47.0
5 2 20.0
6 3 56.0
7 3 30.0
And if all values are ints add astype:
df['b'] = df['b'].fillna(df['a'].map(d)).astype(int)
print (df)
a b
0 1 35
1 1 10
2 1 10
3 2 20
4 2 47
5 2 20
6 3 56
7 3 30
If all data in column a are in keys of dict, then is possible use replace:
df['b'] = df['b'].fillna(df['a'].replace(d))