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How to make a new row by taking the percentage of two other rows in a table on python

So I have a pandas data frame that shows the number of shots taken and the number of goals scored for a list of hockey games from different coordinates. The data frame lists the shots and goals like this (4, 2), and I want to add another column that divides the goals by the shots to give shot percentage for each coordinate. So far here is my code...
key in contents['liveData']['plays']['allPlays']:
# for plays in key['result']['event']:
# print(key)
if (key['result']['event'] == "Shot"):
#print(key['result']['event'])
scoordinates = (key['coordinates']['x'], key['coordinates']['y'])
if scoordinates not in shots:
shots[scoordinates] = (1, 0)
else:
shots[scoordinates] = tuple(map(sum, zip((1, 0), shots[scoordinates])))
if (key['result']['event'] == "Goal"):
#print(key['result']['event'])
gcoordinates = (key['coordinates']['x'], key['coordinates']['y'])
if gcoordinates not in shots:
shots[gcoordinates] = (1, 1)
else:
shots[gcoordinates] = tuple(map(sum, zip((1, 1), shots[gcoordinates])))
#create data frame using pandas
pd.set_option("display.max_rows", None, "display.max_columns", None)
sdf = pd.DataFrame(list(shots.items()),columns = ['Coordinates','Occurences (S, G)'])
file.write(f"{sdf}\n")
this gives the result data frame this--
Coordinates Occurences (S, G)
0 (78.0, -19.0) (2, 1)
1 (-37.0, -10.0) (2, 0)
2 (47.0, -23.0) (3, 1)
3 (53.0, 14.0) (1, 0)
4 (77.0, -2.0) (8, 4)
5 (80.0, 1.0) (12, 5)
6 (74.0, 14.0) (7, 0)
7 (87.0, -3.0) (1, 1)
If anyone can help that would be great!

Try this:
df['new_col']=df['old_col'].apply( lambda x: x[1]/x[0])

Just divide the 2 columns. This is the "longer" way. Split the S, G tuples into their own columns, then divide. Or do the one-liner with lambda that Ave799 provided. Both work, but Ave799 is probably the preferred way
import pandas as pd
data = pd.DataFrame([[(78.0, -19.0),(2, 1)],
[(-37.0, -10.0),(2, 0)],
[(47.0, -23.0),(3, 1)],
[(53.0, 14.0),(1, 0)],
[(77.0, -2.0),(8, 4)],
[(80.0, 1.0),(12, 5)],
[(74.0, 14.0),(7, 0)],
[(87.0, -3.0),(1, 1)]], columns=['Coordinates','Occurences (S, G)'])
data[['S','G']] = pd.DataFrame(data['Occurences (S, G)'].tolist(), index=data.index)
data['Percentage'] = data['G'] / data['S']
Output:
print(data)
Coordinates Occurences (S, G) Percentage S G
0 (78.0, -19.0) (2, 1) 0.500000 2 1
1 (-37.0, -10.0) (2, 0) 0.000000 2 0
2 (47.0, -23.0) (3, 1) 0.333333 3 1
3 (53.0, 14.0) (1, 0) 0.000000 1 0
4 (77.0, -2.0) (8, 4) 0.500000 8 4
5 (80.0, 1.0) (12, 5) 0.416667 12 5
6 (74.0, 14.0) (7, 0) 0.000000 7 0
7 (87.0, -3.0) (1, 1) 1.000000 1 1

Pandas: how to drop rows if contains more that 2 entries?

I have a dataframe like the following
df
entry
0 (5, 4)
1 (4, 2, 1)
2 (0, 1)
3 (2, 7)
4 (9, 4, 3)
I would like to keep only the entry that contains two values
df
entry
0 (5, 4)
1 (0, 1)
2 (1, 7)

If there are tuples use Series.str.len for lengths and compare by Series.le for <= and filter in boolean indexing:
df1 = df[df['entry'].str.len().le(2)]
print (df1)
entry
0 (5, 4)
2 (0, 1)
3 (2, 7)
If there are strings compare number of , and compare by Series.lt for <:
df2 = df[df['entry'].str.count(',').lt(2)]
print (df2)
entry
0 (5,4)
2 (0,1)
3 (2,7)

Repeating the same process for the whole dataset

Given DataFrame df:
1 1.1 2 2.1 ... 1600 1600.1
0 45.1024 7.2365 45.8769 7.1937 34.1072 8.4643
1 43.1024 8.9645 32.5798 7.7500 33.1072 9.3564
2 42.1024 6.7498 25.1027 7.3496 26.1072 6.3665
I did the following operation: I chose first(1 and 1.1) couple and created an array. Then I did the same with following couple (2 and 2.1).
x = df['1']
y = df['1.1']
P = np.array([x, y])
and
q = df['2']
w = df['2.1']
Q = np.array([q, w])
Final operation was:
Q_final = list(zip(Q[0], Q[1]))
P_final = list(zip(P[0], P[1]))
Now I want to do it for the whole dataset. But it will take a lot of time. Any idea how to iterate this in a short way?
EDITED
After all I'm doing
df = similaritymeasures.frechet_dist(P_final, Q_final)
So I want to get a new dataset (maybe) with all columns combinations

A simple way is to use agg across axis 1
def f(s):
s = iter(s)
return list(zip(s,s))
agg = df.agg(f,1)
Then to retrieve, use .str. For example,
agg.str[0] # P_final
agg.str[1] # Q_final
.
.
.
Also, can groupby across axis=1, assuming you want every couple of columns
df.groupby(np.arange(len(df.columns))//2, axis=1).apply(lambda s: s.agg(list,1))

You probably don't want to create 1600 individual variables. Store this in a container, like a dict, where the keys reference the original column handles:
{idx: list(zip(gp.iloc[:, 0], gp.iloc[:, 1]))
for idx, gp in df.groupby(df.columns.str.split('.').str[0], axis=1)}
# or
{idx: [*map(tuple, gp.to_numpy())]
for idx, gp in df.groupby(df.columns.str.split('.').str[0], axis=1)}
Sample
import pandas as pd
import numpy as np
np.random.seed(42)
df = pd.DataFrame((np.random.randint(1,10,(5,6))))
df.columns = ['1', '1.1', '2', '2.1', '3', '3.1']
# 1 1.1 2 2.1 3 3.1
#0 7 4 8 5 7 3
#1 7 8 5 4 8 8
#2 3 6 5 2 8 6
#3 2 5 1 6 9 1
#4 3 7 4 9 3 5
{idx: list(zip(gp.iloc[:, 0], gp.iloc[:, 1]))
for idx, gp in df.groupby(df.columns.str.split('.').str[0], axis=1)}
#{'1': [(7, 4), (7, 8), (3, 6), (2, 5), (3, 7)],
# '2': [(8, 5), (5, 4), (5, 2), (1, 6), (4, 9)],
# '3': [(7, 3), (8, 8), (8, 6), (9, 1), (3, 5)]}

matching two different arrays and making a new array in python

I have two two-dimensional arrays, and I have to create a new array filtering through the 2nd array where 1st column indexes match. The arrays are of different size.
basically the idea is as follow:
file A
#x y
1 2
3 4
2 2
5 4
6 4
7 4
file B
#x1 y1
0 1
1 1
11 1
5 1
7 1
My expected output 2D array should look like
#newx newy
1 1
5 1
7 1
I tried it following way:
match =[]
for i in range(len(x)):
if x[i] == x1[i]:
new_array = x1[i]
match.append(new_array)
print match
This does not seem to work. Please suggest a way to create the new 2D array

Try np.isin.
arr1 = np.array([[1,3,2,5,6,7], [2,4,2,4,4,4]])
arr2 = np.array([[0,1,11,5,7], [1,1,1,1,1]])
arr2[:,np.isin(arr2[0], arr1[0])]
array([[1, 5, 7],
[1, 1, 1]])
np.isin(arr2[0], arr1[0]) checks whether each element of arr2[0] is in arr1[0]. Then, we use the result as the boolean index array to select elements in arr2.

If you make a set out of the first element in A, then it is fairly easy to find the elements in B to keep like:
Code:
a = ((1, 2), (3, 4), (2, 2), (5, 4), (6, 4), (7, 4))
b = ((0, 1), (1, 1), (11, 1), (5, 1), (7, 1))
in_a = {i[0] for i in a}
new_b = [i for i in b if i[0] in in_a]
print(new_b)
Results:
[(1, 1), (5, 1), (7, 1)]
Output results to file as:
with open('output.txt', 'w') as f:
for value in new_b:
f.write(' '.join(str(v) for v in value) + '\n')

#!/usr/bin/env python3
from io import StringIO
import pandas as pd
fileA = """x y
1 2
3 4
2 2
5 4
6 4
7 4
"""
fileB = """x1 y1
0 1
1 1
11 1
5 1
7 1
"""
df1 = pd.read_csv(StringIO(fileA), delim_whitespace=True, index_col="x")
df2 = pd.read_csv(StringIO(fileB), delim_whitespace=True, index_col="x1")
df = pd.merge(df1, df2, left_index=True, right_index=True)
print(df["y1"])
# 1 1
# 5 1
# 7 1
https://pandas.pydata.org/pandas-docs/stable/merging.html#database-style-dataframe-joining-merging

If you use pandas:
import pandas as pd
A = pd.DataFrame({'x': pd.Series([1,3,2,5,6,7]), 'y': pd.Series([2,4,2,4,4,4])})
B = pd.DataFrame({'x1': pd.Series([0,1,11,5,7]), 'y1': 1})
C = A.join(B.set_index('x1'), on='x')
Then if you wanted to drop the unneeded row/columns and rename the columns:
C = A.join(B.set_index('x1'), on='x')
C = C.drop(['y'], axis=1)
C.columns = ['newx', 'newy']
which gives you:
>>> C
newx newy
0 1 1.0
3 5 1.0
5 7 1.0
If you are going to work with arrays, dataframes, etc - pandas is definitely worth a look: https://pandas.pydata.org/pandas-docs/stable/10min.html

Assuming that you have (x, y) pairs in your 2-D arrays, a simple loop may work:
arr1 = [[1, 2], [3, 4], [2, 2]]
arr2 = [[0, 1], [1, 1], [11, 1]]
result = []
for pair1 in arr1:
for pair2 in arr2:
if (pair1[0] == pair2[0]):
result.append(pair2)
print(result)

Not the best solution for smaller arrays, but for really large arrays, works fast -
import numpy as np
import pandas as pd
n1 = np.transpose(np.array([[1,3,2,5,6,7], [2,4,2,4,4,4]]))
n2 = np.transpose(np.array([[0,1,11,5, 7], [1,1,1,1,1]]))
np.array(pd.DataFrame(n1).merge(pd.DataFrame(n2), on=0, how='inner').drop('1_x', axis=1))

Delete duplicate values from pandas DataFrame [duplicate]

Trying to remove duplicate based on unique values on column 'new', I have even tried two methods, but the output df.shape suggests before/after have the same df shape, meaning remove duplication fails.
import pandas
import numpy as np
import random
df = pandas.DataFrame(np.random.randn(10, 4), columns=list('ABCD'))
df['new'] = [1, 1, 3, 4, 5, 1, 7, 8, 1, 10]
df['new2'] = [1, 1, 2, 4, 5, 3, 7, 8, 9, 5]
print df.shape
df.drop_duplicates('new', take_last=False)
df.groupby('new').max()
print df.shape
# output
(10, 6)
(10, 6)
[Finished in 1.0s]

You need to assign the result of drop_duplicates, by default inplace=False so it returns a copy of the modified df, as you don't pass param inplace=True your original df is unmodified:
In [106]:
df = df.drop_duplicates('new', take_last=False)
df.groupby('new').max()
Out[106]:
A B C D new2
new
1 -1.698741 -0.550839 -0.073692 0.618410 1
3 0.519596 1.686003 1.395585 1.298783 2
4 1.557550 1.249577 0.214546 -0.077569 4
5 -0.183454 -0.789351 -0.374092 -1.824240 5
7 -1.176468 0.546904 0.666383 -0.315945 7
8 -1.224640 -0.650131 -0.394125 0.765916 8
10 -1.045131 0.726485 -0.194906 -0.558927 5
if you passed inplace=True it would work:
In [108]:
df.drop_duplicates('new', take_last=False, inplace=True)
df.groupby('new').max()
Out[108]:
A B C D new2
new
1 0.334352 -0.355528 0.098418 -0.464126 1
3 -0.394350 0.662889 -1.012554 -0.004122 2
4 -0.288626 0.839906 1.335405 0.701339 4
5 0.973462 -0.818985 1.020348 -0.306149 5
7 -0.710495 0.580081 0.251572 -0.855066 7
8 -1.524862 -0.323492 -0.292751 1.395512 8
10 -1.164393 0.455825 -0.483537 1.357744 5