Reputation: 187
I have a list of tuples as follows:
data = [
('A', '59', '62'), ('A', '2', '6'), ('A', '87', '92'),
('A', '98', '104'), ('A', '111', '117'),
('B', '66', '71'), ('B', '25', '31'), ('B', '34', '40'), ('B', '46', '53'),
('B', '245', '251'), ('B', '235', '239'), ('B', '224', '229'), ('B', '135', '140'),
('C', '157', '162'), ('C', '203', '208'),
('D', '166', '173'), ('D', '176', '183'),
('E', '59', '62'), ('E', '2', '6'), ('E', '87', '92'), ('E', '98', '104'), ('E', '111', '117')
]
They correspond to a subset of a bigger data-set, so I extracted as above to simplify this post. The first element of each tuple i.e. A, B, C, D, E... is an identifier and can be present in multiple copies.
I would like to extract for each ID/category (A,B,C,D,E...):
1 - minimum from the 2nd element of the tuple
2 - maximum from the 3rd element of the tuple
The final output list should look like:
A: min = 2, max = 117
B: min = 25, max = 251
C: min = 157, max = 208
D: min = 166, max = 183
E: min = 2, max = 117
I tried an approach based on this post: How to remove duplicate from list of tuple when order is important
I simplified for testing by using tuples with only the first 2 elements and extracting the minimum only.
The output looks like this:
('A', '111')
('B', '135')
('C', '157')
('D', '166')
('E', '111')
It should be:
('A', '2')
('B', '25')
('C', '157')
('D', '166')
('E', '2')
I'm looking for an approach that would work with the complete "triple tuple" example, so as to avoid splitting data into multiple subsets.
Many thanks for your time.
2018-10-31
Hello,
please see my edit below that includes the code snippet not included earlier. This gives the erroneous minimum values in the preceding part of the post.
data_min_only = [('A', '59'), ('A', '2'), ('A', '87'), ('A', '98'), ('A', '111'), ('B', '66'), ('B', '25'), ('B', '34'), ('B', '46'), ('B', '245'), ('B', '235'), ('B', '224'), ('B', '135'), ('C', '157'), ('C', '203'), ('D', '166'), ('D', '176'), ('E', '59'), ('E', '2'), ('E', '87'), ('E', '98'), ('E', '111')]
from collections import OrderedDict
empty_dict = OrderedDict()
for item in data_min_only:
# Get old value in dictionary if exist
old = empty_dict.get(item[0])
# Skip if new item is larger than old
if old:
if item[1] > old[1]:
continue
else:
del d[item[0]]
# Assign
empty_dict[item[0]] = item
list(empty_dict.values())
I was thinking that the order of the tuple values for each category was the problem (should be smallest to largest prior to iterating through data_min_only
.
Thank you to all posters for their prompt responses and suggestions/solutions! I'm currently working through those to try and understand and adapt them further.
2018-10-31
I tweaked @slider suggestion to retrieve the differences between min and max. I also tried to output that result to a list as below, but only the last result appears.
for k, g in groupby(sorted(data), key=lambda x: x[0]):
vals = [(int(t[1]), int(t[2])) for t in g]
print (max(i[1] for i in vals) - min(i[0] for i in vals))
test_lst = []
test_lst.append((max(i[1] for i in vals) - min(i[0] for i in vals)))
I also tried this but got the same result:
for i in vals:
test_lst2 = []
test_lst2.append((max(i[1] for i in vals) - min(i[0] for i in vals)))
For this kind of loop, what is the best way to extract the results to a list?
Thanks again.
2018-10-31
test_lst = []
for k, g in groupby(sorted(data), key=lambda x: x[0]):
vals = [(int(t[1]), int(t[2])) for t in g]
print (max(i[1] for i in vals) - min(i[0] for i in vals))
test_lst.append((max(i[1] for i in vals) - min(i[0] for i in vals)))
Solution to extracting loop data - empty list should be outside the loop. Please see @slider comments for his post below.
Upvotes: 2
Views: 5712
Reputation: 2606
This an another approach that will work using the Pandas library:
import pandas as pd
#The same dataset you provided us
data = [('A', '59', '62'), ('A', '2', '6'), ('A', '87', '92'), ('A', '98', '104'), ('A', '111', '117'), ('B', '66', '71'), ('B', '25', '31'), ('B', '34', '40'), ('B', '46', '53'), ('B', '245', '251'), ('B', '235', '239'), ('B', '224', '229'), ('B', '135', '140'), ('C', '157', '162'), ('C', '203', '208'), ('D', '166', '173'), ('D', '176', '183'), ('E', '59', '62'), ('E', '2', '6'), ('E', '87', '92'), ('E', '98', '104'), ('E', '111', '117')]
#Generate dataframe df
df = pd.DataFrame(data=data)
#Convert strings to their respective numerical values
df[[1,2]] = df[[1,2]].apply(pd.to_numeric, errors='ignore')
#Group values using column 0
df.groupby(0).agg({1: min, 2: max})
We use the agg method with a dictionary as the argument in order to find the minimum in column 1 and the maximum in column 2 for each grouped range.
This gives the following result:
1 2
0
A 2 117
B 25 251
C 157 208
D 166 183
E 2 117
Upvotes: 1
Reputation: 12990
You can use itertools.groupby
to first group by the "id" key, and then compute the min and max for each group:
from itertools import groupby
groups = []
for k, g in groupby(sorted(data), key=lambda x: x[0]):
groups.append(list(g))
for g in groups:
print(g[0][0], 'min:', min(int(i[1]) for i in g), 'max:', max(int(i[2]) for i in g))
Output
A min: 2 max: 117
B min: 25 max: 251
C min: 157 max: 208
D min: 166 max: 183
E min: 2 max: 117
Note that you don't have to store the groups first in the groups
list; you can directly print the min and max as you're iterating in the groupby
for loop:
for k, g in groupby(sorted(data), key=lambda x: x[0]):
vals = [(int(t[1]), int(t[2])) for t in g]
print(k, 'min:', min(i[0] for i in vals), 'max:', max(i[1] for i in vals))
Upvotes: 5
Reputation: 4213
data = [('A', '59', '62'), ('A', '2', '6'), ('A', '87', '92'), ('A', '98', '104'), ('A', '111', '117'), ('B', '66', '71'), ('B', '25', '31'), ('B', '34', '40'), ('B', '46', '53'), ('B', '245', '251'), ('B', '235', '239'), ('B', '224', '229'), ('B', '135', '140'), ('C', '157', '162'), ('C', '203', '208'), ('D', '166', '173'), ('D', '176', '183'), ('E', '59', '62'), ('E', '2', '6'), ('E', '87', '92'), ('E', '98', '104'), ('E', '111', '117')]
result = {} # construct result dictionary
for i in data:
cur_min, cur_max = map(int, i[1:])
min_i, max_i = result.setdefault(i[0], [cur_min, cur_max])
if cur_min < min_i:
result[i[0]][0] = cur_min
if cur_max > max_i:
result[i[0]][1] = cur_max
# print(result) # dictionary containing keys with list of min and max values for given key >>> {'A': [2, 117], 'B': [25, 251], 'C': [157, 208], 'D': [166, 183], 'E': [2, 117]}
for k, v in result.items(): # loop to print output
print("{} min: {} max: {}".format(k, v[0], v[1]))
A min: 2 max: 117
B min: 25 max: 251
C min: 157 max: 208
D min: 166 max: 183
E min: 2 max: 117
Upvotes: 2
Reputation: 690
Another approach:
max_list = {}
min_list = {}
for i in data:
if i[0] not in max_list:
max_list[i[0]] = -99999
min_list[i[0]] = 99999
if max_list[i[0]] < int(i[2]):
max_list[i[0]] = int(i[2])
if min_list[i[0]] > int(i[1]):
min_list[i[0]] = int(i[1])
for ele in max_list:
print(ele, ' min: ', min_list[ele], 'max: ', max_list[ele])
Upvotes: 2