I have the following dataframe
import pandas as pd
test = pd.DataFrame({'y':[1,2,3,4,5,6], 'label': ['bottom', 'top','bottom', 'top','bottom', 'top']})
y label
0 1 bottom
1 2 top
2 3 bottom
3 4 top
4 5 bottom
5 6 top
I would like to add a new column, agg_y, which would be the the max(y) if label=="bottom" and min(y) if label=="top". I have tried this
test['min_y'] = test.groupby('label').y.transform('min')
test['max_y'] = test.groupby('label').y.transform('max')
test['agg_y'] = np.where(test.label == "bottom", test.max_y, test.min_y)
test.drop(columns=['min_y', 'max_y'], inplace=True)
which gives the correct result
y label agg_y
0 1 bottom 5
1 2 top 2
2 3 bottom 5
3 4 top 2
4 5 bottom 5
5 6 top 2
I am just looking fora one-liner solution, if possible
Your solution in one line solution is:
test['agg_y'] = np.where(test.label == "bottom",
test.groupby('label').y.transform('max'),
test.groupby('label').y.transform('min'))
Solution without groupby, thank you #ouroboros1:
test['agg_y'] = np.where(test.label == 'bottom',
test.loc[test.label.eq('bottom'), 'y'].max(),
test.loc[test.label.ne('bottom'), 'y'].min())
Another idea is mapping values, idea is similar like ouroboros1 solution:
d = {'bottom':'max', 'top':'min'}
test['agg_y'] = test['label'].map({val:test.loc[test.label.eq(val),'y'].agg(func)
for val, func in d.items()})
print (test)
y label agg_y
0 1 bottom 5
1 2 top 2
2 3 bottom 5
3 4 top 2
4 5 bottom 5
5 6 top 2
Related
I want to do some complex calculations in pandas while referencing previous values (basically I'm calculating row by row). However the loops take forever and I wanted to know if there was a faster way. Everybody keeps mentioning using shift but I don't understand how that would even work.
df = pd.DataFrame(index=range(500)
df["A"]= 2
df["B"]= 5
df["A"][0]= 1
for i in range(len(df):
if i != 0: df['A'][i] = (df['A'][i-1] / 3) - df['B'][i-1] + 25
numpy_ext can be used for expanding calculations
pandas-rolling-apply-using-multiple-columns for reference
I have also included a simpler calc to demonstrate behaviour in simpler way
df = pd.DataFrame(index=range(5000))
df["A"]= 2
df["B"]= 5
df["A"][0]= 1
import numpy_ext as npe
# for i in range(len(df):
# if i != 0: df['A'][i] = (df['A'][i-1] / 3) - df['B'][i-1] + 25
# SO example - function of previous values in A and B
def f(A,B):
r = np.sum(A[:-1]/3) - np.sum(B[:-1] + 25) if len(A)>1 else A[0]
return r
# much simpler example, sum of previous values
def g(A):
return np.sum(A[:-1])
df["AB_combo"] = npe.expanding_apply(f, 1, df["A"].values, df["B"].values)
df["A_running"] = npe.expanding_apply(g, 1, df["A"].values)
print(df.head(10).to_markdown())
sample output
A
B
AB_combo
A_running
0
1
5
1
0
1
2
5
-29.6667
1
2
2
5
-59
3
3
2
5
-88.3333
5
4
2
5
-117.667
7
5
2
5
-147
9
6
2
5
-176.333
11
7
2
5
-205.667
13
8
2
5
-235
15
9
2
5
-264.333
17
I want to create a new column with all the coordinates the car needs to pass to a certain goal. This should be as a list in a panda.
To start with I have this:
import pandas as pd
cars = pd.DataFrame({'x_now': np.repeat(1,5),
'y_now': np.arange(5,0,-1),
'x_1_goal': np.repeat(1,5),
'y_1_goal': np.repeat(10,5)})
output would be:
x_now y_now x_1_goal y_1_goal
0 1 5 1 10
1 1 4 1 10
2 1 3 1 10
3 1 2 1 10
4 1 1 1 10
I have tried to add new columns like this, and it does not work
for xy_index in range(len(cars)):
if cars.at[xy_index, 'x_now'] == cars.at[xy_index,'x_1_goal']:
cars.at[xy_index, 'x_car_move_route'] = np.repeat(cars.at[xy_index, 'x_now'].astype(int),(
abs(cars.at[xy_index, 'y_now'].astype(int)-cars.at[xy_index, 'y_1_goal'].astype(int))))
else:
cars.at[xy_index, 'x_car_move_route'] = \
np.arange(cars.at[xy_index,'x_now'], cars.at[xy_index,'x_1_goal'],
(cars.at[xy_index,'x_1_goal'] - cars.at[xy_index,'x_now']) / (
abs(cars.at[xy_index,'x_1_goal'] - cars.at[xy_index,'x_now'])))
at the end I want the columns x_car_move_route and y_car_move_route so I can loop over the coordinates that they need to pass. I will show it with tkinter. I will also add more goals, since this is actually only the first turn that they need to make.
x_now y_now x_1_goal y_1_goal x_car_move_route y_car_move_route
0 1 5 1 10 [1,1,1,1,1] [6,7,8,9,10]
1 1 4 1 10 [1,1,1,1,1,1] [5,6,7,8,9,10]
2 1 3 1 10 [1,1,1,1,1,1,1] [4,5,6,7,8,9,10]
3 1 2 1 10 [1,1,1,1,1,1,1,1] [3,4,5,6,7,8,9,10]
4 1 1 1 10 [1,1,1,1,1,1,1,1,1] [2,3,4,5,6,7,8,9,10]
You can apply() something like this route() function along axis=1, which means route() will receive rows from cars. It generates either x or y coordinates depending on what's passed into var (from args).
You can tweak/fix as needed, but it should get you started:
def route(row, var):
var2 = 'y' if var == 'x' else 'x'
now, now2 = row[f'{var}_now'], row[f'{var2}_now']
goal, goal2 = row[f'{var}_1_goal'], row[f'{var2}_1_goal']
diff, diff2 = goal - now, goal2 - now2
if diff == 0:
result = np.array([now] * abs(diff2)).astype(int)
else:
result = 1 + np.arange(now, goal, diff / abs(diff)).astype(int)
return result
cars['x_car_move_route'] = cars.apply(route, args=('x',), axis=1)
cars['y_car_move_route'] = cars.apply(route, args=('y',), axis=1)
x_now y_now x_1_goal y_1_goal x_car_move_route y_car_move_route
0 1 5 1 10 [1,1,1,1,1] [6,7,8,9,10]
1 1 4 1 10 [1,1,1,1,1,1] [5,6,7,8,9,10]
2 1 3 1 10 [1,1,1,1,1,1,1] [4,5,6,7,8,9,10]
3 1 2 1 10 [1,1,1,1,1,1,1,1] [3,4,5,6,7,8,9,10]
4 1 1 1 10 [1,1,1,1,1,1,1,1,1] [2,3,4,5,6,7,8,9,10]
I am new to python and the last time I coded was in the mid-80's so I appreciate your patient help.
It seems .rolling(window) requires the window to be a fixed integer. I need a rolling window where the window or lookback period is dynamic and given by another column.
In the table below, I seek the Lookbacksum which is the rolling sum of Data as specified by the Lookback column.
d={'Data':[1,1,1,2,3,2,3,2,1,2],
'Lookback':[0,1,2,2,1,3,3,2,3,1],
'LookbackSum':[1,2,3,4,5,8,10,7,8,3]}
df=pd.DataFrame(data=d)
eg:
Data Lookback LookbackSum
0 1 0 1
1 1 1 2
2 1 2 3
3 2 2 4
4 3 1 5
5 2 3 8
6 3 3 10
7 2 2 7
8 1 3 8
9 2 1 3
You can create a custom function for use with df.apply, eg:
def lookback_window(row, values, lookback, method='sum', *args, **kwargs):
loc = values.index.get_loc(row.name)
lb = lookback.loc[row.name]
return getattr(values.iloc[loc - lb: loc + 1], method)(*args, **kwargs)
Then use it as:
df['new_col'] = df.apply(lookback_window, values=df['Data'], lookback=df['Lookback'], axis=1)
There may be some corner cases but as long as your indices align and are unique - it should fulfil what you're trying to do.
here is one with a list comprehension which stores the index and value of the column df['Lookback'] and the gets the slice by reversing the values and slicing according to the column value:
df['LookbackSum'] = [sum(df.loc[:e,'Data'][::-1].to_numpy()[:i+1])
for e,i in enumerate(df['Lookback'])]
print(df)
Data Lookback LookbackSum
0 1 0 1
1 1 1 2
2 1 2 3
3 2 2 4
4 3 1 5
5 2 3 8
6 3 3 10
7 2 2 7
8 1 3 8
9 2 1 3
An exercise in pain, if you want to try an almost fully vectorized approach. Sidenote: I don't think it's worth it here. At all.
Inspired by Divakar's answer here
Given:
import numpy as np
import pandas as pd
d={'Data':[1,1,1,2,3,2,3,2,1,2],
'Lookback':[0,1,2,2,1,3,3,2,3,1],
'LookbackSum':[1,2,3,4,5,8,10,7,8,3]}
df=pd.DataFrame(data=d)
Using the function from Divakar's answer, but slightly modified
from skimage.util.shape import view_as_windows as viewW
def strided_indexing_roll(a, r, fill_value=np.nan):
# Concatenate with sliced to cover all rolls
p = np.full((a.shape[0],a.shape[1]-1),fill_value)
a_ext = np.concatenate((p,a,p),axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = a.shape[1]
return viewW(a_ext,(1,n))[np.arange(len(r)), -r + (n-1),0]
Now, we just need to prepare a 2d array for the data and independently shift the rows according to our desired lookback values.
arr = df['Data'].to_numpy().reshape(1, -1).repeat(len(df), axis=0)
shifter = np.arange(len(df) - 1, -1, -1) #+ d['Lookback'] - 1
temp = strided_indexing_roll(arr, shifter, fill_value=0)
out = strided_indexing_roll(temp, (len(df) - 1 - df['Lookback'])*-1, 0).sum(-1)
Output:
array([ 1, 2, 3, 4, 5, 8, 10, 7, 8, 3], dtype=int64)
We can then just assign it back to the dataframe as needed and check.
df['out'] = out
#output:
Data Lookback LookbackSum out
0 1 0 1 1
1 1 1 2 2
2 1 2 3 3
3 2 2 4 4
4 3 1 5 5
5 2 3 8 8
6 3 3 10 10
7 2 2 7 7
8 1 3 8 8
9 2 1 3 3
I am a beginner and this is my first project.. I searched for the answer but it still isn't clear.
I have imported a worksheet from excel using Pandas..
**Rabbit Class:
Num Behavior Speaking Listening
0 1 3 1 1
1 2 1 1 1
2 3 3 1 1
3 4 1 1 1
4 5 3 2 2
5 6 3 2 3
6 7 3 3 1
7 8 3 3 3
8 9 2 3 2
What I want to do is create if functions.. ex. if a student's behavior is a "1" I want it to print one string, else print a different string. How can I reference a particular cell of the worksheet to set up such a function? I tried: val = df.at(1, "Behavior") but that clearly isn't working..
Here is the code I have so far..
import os
import pandas as pd
from pandas import ExcelWriter
from pandas import ExcelFile
path = r"C:\Users\USER\Desktop\Python\rabbit_class.xls"
print("Rabbit Class:")
print(df)
Also you can do
dff = df.loc[df['Behavior']==1]
if(not(dff.empty)):
# do Something
What you want is to find rows where df.Behavior is equal to 1. Use any of the following three methods.
# Method-1
df[df["Behavior"]==1]
# Method-2
df.loc[df["Behavior"]==1]
# Method-3
df.query("Behavior==1")
Output:
Num Behavior Speaking Listening LastColumn
0 0 1 3 1 1
Note: Dummy Data
Your sample data does not have a column header (the last one). So I named it LastColumn and read-in the data as a dataframe.
# Dummy Data
s = """
Num Behavior Speaking Listening LastColumn
0 1 3 1 1
1 2 1 1 1
2 3 3 1 1
3 4 1 1 1
4 5 3 2 2
5 6 3 2 3
6 7 3 3 1
7 8 3 3 3
8 9 2 3 2
"""
# Make Dataframe
ss = re.sub('\s+',',',s)
ss = ss[1:-1]
sa = np.array(ss.split(',')).reshape(-1,5)
df = pd.DataFrame(dict((k,v) for k,v in zip(sa[0,:], sa[1:,].T)))
df = df.astype(int)
df
Hope below example will help you
import pandas as pd
df = pd.read_excel(r"D:\test_stackoverflow.xlsx")
print(df.columns)
def _filter(col, filter_):
return df[df[col]==filter_]
print(_filter('Behavior', 1))
Thank you all for your answers. I finally figured out what I was trying to do using the following code:
i = 0
for i in df.index:
student_number = df["Student Number"][i]
print(student_number)
student_name = student_list[int(student_number) - 1]
behavior = df["Behavior"][i]
if behavior == 1:
print("%s's behavior is good" % student_name)
elif behavior == 2:
print ("%s's behavior is average." % student_name)
else:
print ("%s's behavior is poor" % student_name)
speaking = df["Speaking"][i]
I want to draw bar chart for below data:
4 1406575305 4
4 -220936570 2
4 2127249516 2
5 -1047108451 4
5 767099153 2
5 1980251728 2
5 -2015783241 2
6 -402215764 2
7 927697904 2
7 -631487113 2
7 329714360 2
7 1905727440 2
8 1417432814 2
8 1906874956 2
8 -1959144411 2
9 859830686 2
9 -1575740934 2
9 -1492701645 2
9 -539934491 2
9 -756482330 2
10 1273377106 2
10 -540812264 2
10 318171673 2
The 1st column is the x-axis and the 3rd column is for y-axis. Multiple data exist for same x-axis value. For example,
4 1406575305 4
4 -220936570 2
4 2127249516 2
This means three bars for 4 value of x-axis and each of bar is labelled with tag(the value in middle column). The sample bar chart is like:
http://matplotlib.org/examples/pylab_examples/barchart_demo.html
I am using matplotlib.pyplot and np. Thanks..
I followed the tutorial you linked to, but it's a bit tricky to shift them by a nonuniform amount:
import numpy as np
import matplotlib.pyplot as plt
x, label, y = np.genfromtxt('tmp.txt', dtype=int, unpack=True)
ux, uidx, uinv = np.unique(x, return_index=True, return_inverse=True)
max_width = np.bincount(x).max()
bar_width = 1/(max_width + 0.5)
locs = x.astype(float)
shifted = []
for i in range(max_width):
where = np.setdiff1d(uidx + i, shifted)
locs[where[where<len(locs)]] += i*bar_width
shifted = np.concatenate([shifted, where])
plt.bar(locs, y, bar_width)
If you want you can label them with the second column instead of x:
plt.xticks(locs + bar_width/2, label, rotation=-90)
I'll leave doing both of them as an exercise to the reader (mainly because I have no idea how you want them to show up).