How to compare particular element in list python3? - python-3.x

l1= [['1', 'apple', '1', '2', '1', '0', '0', '0'], ['1',
'cherry', '1', '1', '1', '0', '0', '0']]
l2 = [['1', 'cherry', '2', '1'],
['1', 'plums', '2', '15'],
['1', 'orange', '2', '15'],
['1', 'cherry', '2', '1'],
['1', 'cherry', '2', '1']]
output = []
for i in l1:
for j in l2:
if i[1] != j[1]:
output.append(j)
break
print(output)
Expected Output:
[['1', 'plums', '2', '15'], ['1', 'orange', '2', '15']]
How to stop iteration and find unique elements and get the sublist?
How to stop iteration and find unique elements and get the sublist?


To find the elements in L2 that are not in L1 based on the fruit name:
l1= [[1,'apple',3],[1,'cherry',4]]
l2 = [[1,'apple',3],[1,'plums',4],[1,'orange',3],[1,'apple',4]]
output = []
for e in l2:
if not e[1] in [f[1] for f in l1]: # search by matching fruit
output.append(e)
print(output)
Output
[[1, 'plums', 4], [1, 'orange', 3]]


You can store all the unique elements from list1 in a new list, then check for list2 if that element exists in the new list. Something like:
newlist = []
for item in l1:
if item[1] not in newlist:
newlist.append(item)
output = []
for item in l2:
if item[1] not in newlist:
output.append(item)
print(output)
This is slightly inefficient but really straightforward to understand.

Related

Combing lists into list in Python

I have these lists in Python
text_0 = ['Weight','Weight', 'Weight']
text_1 = [['x','y','z'],['x','y','z'],['x','y','z']]
text_2 = [['1','2,','3'],['4','5','6'],['7','8','9']]
I would like would to create a new list like this
new_list = ['Weight','x','1','y','2','z','3','Weight','x','4','y','5','z','6','Weight','x','7','y','8','z','9']
How do I do this in Python?

In your case we can do with python
sum([[x] + [ items for item in zip(y,z) for items in item]
for x, y, z in zip(text_0,text_1,text_2)],[])
['Weight', 'x', '1', 'y', '2', 'z', '3', 'Weight', 'x', '4', 'y', '5', 'z', '6', 'Weight', 'x', '7', 'y', '8', 'z', '9']

Adding a string to an array adds all characters separately

Any suggestion how to merge it better so the dual digits numbers does not split?
Sorry for bad english.
def merge(strArr):
newList = []
for x in range(len(strArr)):
newList += strArr[x]
return newList
array_test = ["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"]
print(merge(array_test))
output =['1', ',', ' ', '3', ',', ' ', '4', ',', ' ', '7', ',', ' ', '1', '3', '1', ',', ' ', '2', ',', ' ', '4', ',', ' ', '1', '3', ',', ' ', '1', '5']`
expected output= [1,2,3,4,7,13,1,2,4,13,15]

Using list comprehension:
merged_arr = [n for s in array_test for n in s.split(", ")]
print(merged_arr)
This prints:
['1', '3', '4', '7', '13', '1', '2', '4', '13', '15']

It merges this way because for lists += is an array concatenation and that in this context your string object is interpreted as an array of characters:
[] += "Hello"
# Equivalent to
[] += ["H", "e", "l", "l", "o"]
If you want to join strings you can do:
out = "".join(array_test)

Your result becomes the way it is, because you take each inner string and add each character of it to your return-list without removing any spaces or commas.
You can change your code to:
def merge(strArr):
new_list = []
for inner in strArr:
new_list.extend(inner.split(", ")) # split and extend instead of += (== append)
return new_list
array_test =["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"]
merged = merge(array_test)
as_int = list(map(int,merged))
print(merged)
print(as_int)
Output:
['1', '3', '4', '7', '13', '1', '2', '4', '13', '15']
[1, 3, 4, 7, 13, 1, 2, 4, 13, 15]
Without as_int()you will still hav strings in your list, you need to convert them into integers.

Convert a list of strings to a list of strings and ints

I have a dictionary of lists, each list has 3 elements, i want to convert the 2nd and 3rd element to ints from strings
dict = {1:['string', '2', '3',],
2:['string', '2', '3',],
3:['string', '2', '3',],}
to become:
dict = {1:['string', 2, 3,],
2:['string', 2, 3,],
3:['string', 2, 3,],}
Thank you

Firstly, don't name your dictionaries dict, as it's a reserved keyword and you don't wanna overwrite it.
Coming to your solution.
d = {
1: ['string', '23', '3'],
2: ['string', '2', '3'],
3: ['string', '2', '3'],
}
d2 = {
k: [
int(i) if i.isdigit() else i
for i in v
]
for k, v in d.items()
}
Will give you an output of:
{
1: ['string', '23', '3'],
2: ['string', '2', '3'],
3: ['string', '2', '3']
}
{
1: ['string', 23, 3],
2: ['string', 2, 3],
3: ['string', 2, 3]
}

If you dictionary has many elements you might not want to create a second dictionary in memory, but modify the existing one in-place:
data = {
1: ['string', '2', '3'],
2: ['string', '2', '3'],
3: ['string', '2', '3'],
}
for v in data.values():
v.append(int(v.pop(1)))
v.append(int(v.pop(1)))
print(data)
Output:
{1: ['string', 2, 3], 2: ['string', 2, 3], 3: ['string', 2, 3]}

remove multiple rows from a array in python

array([
['192', '895'],
['14', '269'],
['1', '23'],
['1', '23'],
['50', '322'],
['19', '121'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['2,546', '17,610'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['151', '1,150'],
['45', '319'],
['44', '335'],
['30', '184']
])
How can I remove some of the rows and left the array like:
Table3=array([
['192', '895'],
['14', '269'],
['1', '23'],
['50', '322'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['45', '319'],
['44', '335'],
['30', '184']
])
I removed the index 2,4,6. I am not sure how should I do it. I have tried few ways, but still can't work.

It seems like you actually deleted indices 2, 5, and 10 (not 2, 4 and 6). To do this you can use np.delete, pass it a list of the indices you want to delete, and apply it along axis=0:
Table3 = np.delete(arr, [[2,5,10]], axis=0)
>>> Table3
array([['192', '895'],
['14', '269'],
['1', '23'],
['50', '322'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['151', '1,150'],
['45', '319'],
['44', '335'],
['30', '184']],
dtype='<U6')

How to make list of tuple from list of list?

How do I convert this list of lists:
[['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
To this list of tuples:
[(0, [1, 2, 3]), (1, [0, 4, 6, 7, 9]), (2, [0, 3, 6, 8, 9])]
I am unsure how to implement this next step? (I can't use dictionaries,
sets, deque, bisect module. You can though, and in fact should, use .sort or sorted functions.)
Here is my attempt:
network= [['10'], ['0 1'], ['0 2'], ['0 3'], ['1 4'], ['1 6'], ['1 7'], ['1 9'], ['2 3'], ['2 6'], ['2 8'], ['2 9']]
network.remove(network[0])
friends=[]
for i in range(len(network)):
element= (network[i][0]).split(' ')
friends.append(element)
t=len(friends)
s= len(friends[0])
lst=[]
for i in range(t):
a= (friends[i][0])
if a not in lst:
lst.append(int(a))
for i in range(t):
if a == friends[i][0]:
b=(friends[i][1])
lst.append([b])
print(tuple(lst))
It outputs:
(0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'])
I am very close it seems, not sure what to do??

A simpler method:
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=set(i[0] for i in l)
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)
Output:
[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]
Alternate Answer (without using set)
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=[]
for i in l:
if i[0] not in a:
a.append(i[0])
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)
also outputs
[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]

You can use Pandas:
import pandas as pd
import numpy as np
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
df = pd.DataFrame(l, dtype=np.int)
s = df.groupby(0)[1].apply(list)
list(zip(s.index, s))
Output:
[(0, [1, 2, 3]), (1, [4, 6, 7, 9]), (2, [3, 6, 8, 9])]

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