I have a dataframe df as follows:
Col1 Price Day
A 16 5
B 12 3
D 5 8
I need to apply a function to each row of df:
import pandas as pd, numpy as np
def Fn(Price, Day):
pr = np.arange(Price/2, Price + Price/2, Price/2)
da = np.arange(Day/2, Day+ Day/2, Day/2)
return pd.DataFrame({'Price':pr, 'Day':da)
I need to achieve the following:
Col1 Price Day
A 8 2.5
A 16 5
B 6 1.5
B 12 3
D 2.5 4
D 5 8
In reality with the function Fn has something like:
pr = np.arange(Price/18, Price + Price/18, Price/18)
da = np.arange(Day/18, Day+ Day/18, Day/18)
I am not sure how to proceed with the above.
A possible solution, which:
Iterates over the rows of the dataframe with map
Applies Fn in each iteration, getting the corresponding resulting dataframe, which is put into a list.
Finally, concatenates all dataframes of the mentioned list into a single dataframe.
(pd.concat(map(
lambda x: pd.concat(
[pd.Series(x[1]['Col1'], name='Col1'),
Fn(x[1]['Price'], x[1]['Day'])], axis=1, ignore_index=True),
df.iterrows()))
.ffill()
.set_axis(df.columns, axis=1))
Output:
Col1 Price Day
0 A 8.0 2.5
1 A 16.0 5.0
0 B 6.0 1.5
1 B 12.0 3.0
0 D 2.5 4.0
1 D 5.0 8.0
Related
My table has 4 columns: order_id, item_id_1, item_id_2 and item_id_3. The three last columns cover the same type of information (the ids of products). I want to transform this table to get 2-columns table with "order_id" and "item_id", so my columns cover unique type of informations. That means, if in a particular order_id there were 3 products ordered, I will get three (instead of one) rows in my new table).
This will alow me, for exapmle, perform 'grupby' operation on 'item_id" column to count how meny times a particular product was ordered.
How this table transformation process is called?
For example, if you have a dataframe like this -
df = pd.DataFrame({'order_id':[1,2,3], 'item_id_1':['a','b','c'], 'item_id_2':['x','y',''], 'item_id_3':['','q','']})
df
order_id item_id_1 item_id_2 item_id_3
0 1 a x
1 2 b y q
2 3 c
pd.melt(df, id_vars=['order_id'], \
value_vars=['item_id_1', 'item_id_2', 'item_id_3'], \
var_name='item_id', value_name='item_value').\
replace('',np.nan).dropna().\
sort_values(['order_id']).\
reset_index(drop=True)\
[['order_id', 'item_id']]
So I'm not aware of any method that allows you to expand rows automatically as you're suggesting, but you can easily reach you're goal without. Let's start from a similar data frame, I put nan in cells of items that have not been ordered:
import pandas as pd
import numpy as np
data = {'order_id':[1,2,3],'item_id_1':[11,12,13],'item_id_2':[21,np.nan,23],'item_id_3':[31,np.nan,np.nan]}
df = pd.DataFrame(data)
cols = ['item_id_1','item_id_2','item_id_3']
print(df)
Out:
order_id item_id_1 item_id_2 item_id_3
0 1 11 21.0 31.0
1 2 12 NaN NaN
2 3 13 23.0 NaN
Then you can define a new empty data frame to fill by iterating through the rows of the initial one. For every item a new row is added to the empty data frame with same order_id and different item_id.
new_df = pd.DataFrame(columns = ['order_id','item_id']) # ,'item_num']
for ind, row in df.iterrows():
new_row = {}
new_row['order_id'] = row['order_id']
for col in cols: # for num, col in enumerate(cols):
item = row[col]
if not pd.isna(item):
new_row['item_id'] = item
# new_row['item_num'] = num +1
new_df = new_df.append(new_row,ignore_index=True)
print(new_df)
Out: # shape (6,2), ok because because 6 items have been ordered
order_id item_id
0 1.0 11.0
1 1.0 21.0
2 1.0 31.0
3 2.0 12.0
4 3.0 13.0
5 3.0 23.0
If you want, you could also add a third column to keep trace of the category of each item (i.e. if it was item_1, 2 or 3) by uncommenting the lines in the code, which gives you this output:
order_id item_id item_num
0 1.0 11.0 1.0
1 1.0 21.0 2.0
2 1.0 31.0 3.0
3 2.0 12.0 1.0
4 3.0 13.0 1.0
5 3.0 23.0 2.0
I have a pandas dataframe
x
1
3
4
7
10
I want to create a new column y as y[i] = x[i] - x[i-1] (and y[0] = x[0]).
So the above data frame will become:
x y
1 1
3 2
4 1
7 3
10 3
How to do that with python-3? Many thanks
Using .shift() and fillna():
df['y'] = (df['x'] - df['x'].shift(1)).fillna(df['x'])
To explain what this is doing, if we print(df['x'].shift(1)) we get the following series:
0 NaN
1 1.0
2 3.0
3 4.0
4 7.0
Which is your values from 'x' shifted down one row. The first row gets NaN because there is no value above it to shift down. So, when we do:
print(df['x'] - df['x'].shift(1))
We get:
0 NaN
1 2.0
2 1.0
3 3.0
4 3.0
Which is your subtracted values, but in our first row we get a NaN again. To clear this, we use .fillna(), telling it that we want to just take the value from df['x'] whenever a null value is encountered.
I am new to Python and Pandas , can someone help me with below report.
I want to report difference of N columns and create new columns with difference value, is it possible to make it dynamic as I have more than 30 columns. (Columns are fixed numbers, rows values can change)
A and B can be Alpha numeric
Use join with sub for difference of DataFrames:
#if columns are strings, first cast it
df1 = df1.astype(int)
df2 = df2.astype(int)
#if first columns are not indices
#df1 = df1.set_index('ID')
#df2 = df2.set_index('ID')
df = df1.join(df2.sub(df1).add_prefix('sum'))
print (df)
A B sumA sumB
ID
0 10 2.0 5 3.0
1 11 3.0 6 5.0
2 12 4.0 7 5.0
Or similar:
df = df1.join(df2.sub(df1), rsuffix='sum')
print (df)
A B Asum Bsum
ID
0 10 2.0 5 3.0
1 11 3.0 6 5.0
2 12 4.0 7 5.0
Detail:
print (df2.sub(df1))
A B
ID
0 5 3.0
1 6 5.0
2 7 5.0
IIUC
df1[['C','D']]=(df2-df1)[['A','B']]
df1
Out[868]:
ID A B C D
0 0 10 2.0 5 3.0
1 1 11 3.0 6 5.0
2 2 12 4.0 7 5.0
df1.assign(B=0)
Out[869]:
ID A B C D
0 0 10 0 5 3.0
1 1 11 0 6 5.0
2 2 12 0 7 5.0
The 'ID' column should really be an index. See the Pandas tutorial on indexing for why this is a good idea.
df1 = df1.set_index('ID')
df2 = df2.set_index('ID')
df = df1.copy()
df[['C', 'D']] = df2 - df1
df['B'] = 0
print(df)
outputs
A B C D
ID
0 10 0 5 3.0
1 11 0 6 5.0
2 12 0 7 5.0
Given the following data frame:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Site':['a','a','a','b','b','b'],
'x':[1,1,0,1,0,0],
'y':[1,np.nan,0,1,1,0]
})
df
Site y x
0 a 1.0 1
1 a NaN 1
2 a 0.0 0
3 b 1.0 1
4 b 1.0 0
5 b 0.0 0
I am looking for the most efficient way, for each numerical column (y and x), to produce a percent per group, label the column name, and stack them in one column.
Here's how I accomplish this for 'y':
df=df.loc[~np.isnan(df['y'])] #do not count non-numbers
t=pd.pivot_table(df,index='Site',values='y',aggfunc=[np.sum,len])
t['Item']='y'
t['Perc']=round(t['sum']/t['len']*100,1)
t
sum len Item Perc
Site
a 1.0 2.0 y 50.0
b 2.0 3.0 y 66.7
Now all I need is a way to add 2 more rows to this; the results for 'x' if I had pivoted with its values above, like this:
sum len Item Perc
Site
a 1.0 2.0 y 50.0
b 2.0 3.0 y 66.7
a 1 2 x 50.0
b 1 3 x 33.3
In reality, I have 48 such numerical data columns that need to be stacked as such.
Thanks in advance!
First you can use notnull. Then omit in pivot_table parameter value, stack and sort_values by new column Item. Last you can use pandas function round:
df=df.loc[df['y'].notnull()]
t=pd.pivot_table(df,index='Site', aggfunc=[sum,len])
.stack()
.reset_index(level=1)
.rename(columns={'level_1':'Item'})
.sort_values('Item', ascending=False)
t['Perc']= (t['sum']/t['len']*100).round(1)
#reorder columns
t = t[['sum','len','Item','Perc']]
print t
sum len Item Perc
Site
a 1.0 2.0 y 50.0
b 2.0 3.0 y 66.7
a 1.0 2.0 x 50.0
b 1.0 3.0 x 33.3
Another solution if is neccessary define values columns in pivot_table:
df=df.loc[df['y'].notnull()]
t=pd.pivot_table(df,index='Site',values=['y', 'x'], aggfunc=[sum,len])
.stack()
.reset_index(level=1)
.rename(columns={'level_1':'Item'})
.sort_values('Item', ascending=False)
t['Perc']= (t['sum']/t['len']*100).round(1)
#reorder columns
t = t[['sum','len','Item','Perc']]
print t
sum len Item Perc
Site
a 1.0 2.0 y 50.0
b 2.0 3.0 y 66.7
a 1.0 2.0 x 50.0
b 1.0 3.0 x 33.3
suppose I have two dataframes:
import pandas
....
....
test1 = pandas.DataFrame([1,2,3,4,5])
....
....
test2 = pandas.DataFrame([4,2,1,3,7])
....
I tried test1.append(test2) but it is the equivalent of R's rbind.
How can I combine the two as two columns of a dataframe similar to the cbind function in R?
test3 = pd.concat([test1, test2], axis=1)
test3.columns = ['a','b']
(But see the detailed answer by #feng-mai, below)
There is a key difference between concat(axis = 1) in pandas and cbind() in R:
concat attempts to merge/align by index. There is no concept of index in a R dataframe. If the two pandas dataframes' indexes are misaligned, the results are different from cbind (even if they have the same number of rows). You need to either make sure the indexes align or drop/reset the indexes.
Example:
import pandas as pd
test1 = pd.DataFrame([1,2,3,4,5])
test1.index = ['a','b','c','d','e']
test2 = pd.DataFrame([4,2,1,3,7])
test2.index = ['d','e','f','g','h']
pd.concat([test1, test2], axis=1)
0 0
a 1.0 NaN
b 2.0 NaN
c 3.0 NaN
d 4.0 4.0
e 5.0 2.0
f NaN 1.0
g NaN 3.0
h NaN 7.0
pd.concat([test1.reset_index(drop=True), test2.reset_index(drop=True)], axis=1)
0 1
0 1 4
1 2 2
2 3 1
3 4 3
4 5 7
pd.concat([test1.reset_index(), test2.reset_index(drop=True)], axis=1)
index 0 0
0 a 1 4
1 b 2 2
2 c 3 1
3 d 4 3
4 e 5 7