First off i am certain that such a basic thing has been asked before, but i could not find a post about it.
I have this piece of example data:
'192.168.244.213': ['8', '4', '3', '1', '6', '5', '3', '2', '6', '5'],
'192.168.244.214': ['6', '8', '7', '6', '5', '4', '2', '7', '5', '5'],
'192.168.244.215': ['4', '10', '0', '8', '7', '0', '4', '3', '2', '6'],
'192.168.244.230': ['0', '0', '0', '0', '0', '0', '0', '0', '0', '0']
And i want to print out every line (each line is one dictionary key-value pair) that has a list-value whose list contains any amount of items that is not 0 (in this case, every line except the 4th)
I just cant seem to figure out this seemingly simple thing - what i tried before was those two things:
for i in d.keys():
if "0" not in d[i]:
print(i, d[i])
This one shows only lists that do not contain 0 AT ALL - so the third line would not be shown, even though it contains non-0 values
for i in d.keys():
for j in d[i]:
if j is not "0":
print(i, d[i])
This one DOES show me what i want, but as you can tell, it prints every result way too often - one print for every list value that is not 0.
You can simply iterate over like
def all_zero(arr):
for i in arr:
if i != 0:
return False
else:
return True
You can call it on all the lists one by one.
Use a dictionary-comprehension:
d = {'192.168.244.213': ['8', '4', '3', '1', '6', '5', '3', '2', '6', '5'],
'192.168.244.214': ['6', '8', '7', '6', '5', '4', '2', '7', '5', '5'],
'192.168.244.215': ['4', '10', '0', '8', '7', '0', '4', '3', '2', '6'],
'192.168.244.230': ['0', '0', '0', '0', '0', '0', '0', '0', '0', '0']}
result = {k: v for k, v in d.items() if not all(x == '0' for x in v)}
# {'192.168.244.213': ['8', '4', '3', '1', '6', '5', '3', '2', '6', '5'],
# '192.168.244.214': ['6', '8', '7', '6', '5', '4', '2', '7', '5', '5'],
# '192.168.244.215': ['4', '10', '0', '8', '7', '0', '4', '3', '2', '6']}
The above code generates a new dictionary which omits all items where values are all zeros.
Now that you have a dictionary, you can easily do an iteration like so:
for k, v in result.items():
print(k, v)
Your bug is basically just a missing break:
for i in d.keys():
for j in d[i]:
if j != "0":
print(i, d[i])
break
However, for conciseness I would recommend you check out the any() function, which does exactly what you want: Return true if any of the elements of the iterable are true (when cast to booleans).
Eg:
for i in d.keys():
if any(j != "0" for j in d[i]):
print(i, d[i])
(The j is not "0" generator is only necessary because you have string values. For an int array, any(d[i]) would work.)
Even more "Pythonic" would be removing the need for a dictionary lookup:
for i, d_i in d.items():
if any(j != "0" for j in d_i):
print(i, d_i)
I like the other answers but I feel like you can get away with something like this as well:
for i in d.keys():
#If there are as many zeroes as there are elements in the list...
if d[i].count(0) == len(d[i]):
#...You might as well skip it :)
continue
print(d[i])
for i in d.keys():
all_zero = True
for j in d[i]:
if j is not "0":
all_zero = False
break
if not all_zero:
print(i, d[i])
This may work for almost every language :)
Have a look at how I could accomplish this.
d = {
'192.168.244.213': ['8', '4', '3', '1', '6', '5', '3', '2', '6', '5'],
'192.168.244.214': ['6', '8', '7', '6', '5', '4', '2', '7', '5', '5'],
'192.168.244.215': ['4', '10', '0', '8', '7', '0', '4', '3', '2', '6'],
'192.168.244.230': ['0', '0', '0', '0', '0', '0', '0', '0', '0', '0']
}
for key in d.keys():
if all( item == '0' for item in d[key]):
pass
else:
print(key, d[key])
You should use all in this case, consider following example:
digits = ['0', '2', '0', '4', '7', '5', '0', '3', '2', '6']
zeros = ['0', '0', '0', '0', '0', '0', '0', '0', '0', '0']
print(all([k=='0' for k in digits])) #gives False
print(all([k=='0' for k in zeros])) #gives True
Please remember to deliver [k=='0' for k in ...] to all, as delivering list directly would give True for both digits and zeros, as both contain at least one non-empty str (str of length 1 or greater).
Related
I have a dictionary say : my_map_dict which looks something like below:
{2: [['1', '2', '4', '4', '0', '2', '0', '0.67'], ['5', '6', '3', '8', '0', '2', '1', '0.67'], ['6', '9', '4', '9', '0', '2', '2', '0.67'], ['4', '3', '6', '9', '0', '2', '3', '1.00']], 3: [['4', '6', '6', '1', '0', '3', '0', '0.67'], ['5', '9', '4', '8', '0', '3', '1', '0.67']], 4: [['1', '9', '4', '9', '0', '4', '0', '0.67']]}
Where 2,3,4 are the keys of the dictionary and those lists are the values of that.
Now I want to generate a file from this say outputfile.txt which will take all the values of the dictionary and make that a CSV file which looks like below (desired output):
1,2,4,4,0,2,0,0.67
5,6,3,8,0,2,1,0.67
6,9,4,9,0,2,2,0.67
4,3,6,9,0,2,3,1.00
4,6,6,1,0,3,0,0.67
5,9,4,8,0,3,1,0.67
1,9,4,9,0,4,0,0.67
How can I do it?
Flatten the values of your dictionary to a single list of lists, and use the csv module to write to file:
import csv
with open("output.csv","w",newline="") as f:
writer = csv.writer(f)
writer.writerows([item for sublist in d.values() for item in sublist])
output.csv
1,2,4,4,0,2,0,0.67
5,6,3,8,0,2,1,0.67
6,9,4,9,0,2,2,0.67
4,3,6,9,0,2,3,1.00
4,6,6,1,0,3,0,0.67
5,9,4,8,0,3,1,0.67
1,9,4,9,0,4,0,0.67
I am trying to implement insertion sort using python. Below is my code:
import numpy as np
N=int(input())
A=input().split()
#A=[int(x) for x in input().split()]
B_sorted=A.copy()
C= A.copy()
def inssort(arr):
for i in range(1, len(arr)):
key = arr[i]
j = i-1
while (j >=0 and key < arr[j]):
arr[j+1] = arr[j]
j -= 1
arr[j+1] = key
inssort(B_sorted)
for i in range(0,len(A)):
for j in range(0,len(A)):
if A[i]==B_sorted[j]:
C[i]=j+1
break
j=j+1
i=i+1
for x in B_sorted:
print (x, "",end="")
My code works for the following set of entries:
[9,8,7,...,2,1]
[20,19,...,12,11]
[20,19,...,11,10]
but it doesn't work for following
[10,9,8,...,2,1]
I don't understand what exactly is the issue here.
I tried to go to the detail by printing the output at every loop:
10 9 8 7 6 5 4 3 2 1
['10', '9', '8', '7', '6', '5', '4', '3', '2', '1']
['10', '8', '9', '7', '6', '5', '4', '3', '2', '1']
['10', '7', '8', '9', '6', '5', '4', '3', '2', '1']
['10', '6', '7', '8', '9', '5', '4', '3', '2', '1']
['10', '5', '6', '7', '8', '9', '4', '3', '2', '1']
['10', '4', '5', '6', '7', '8', '9', '3', '2', '1']
['10', '3', '4', '5', '6', '7', '8', '9', '2', '1']
['10', '2', '3', '4', '5', '6', '7', '8', '9', '1']
['1', '10', '2', '3', '4', '5', '6', '7', '8', '9']
Looks like for some reason at the first iteration j is not taking the value 0 for this particular sequence. As I mentioned before, it works perfectly for a lot of other sequences.
Kindly help.
input = '1+2++3+++4++5+6+7++8+9++10'
string = input.split('+')
print(string)
when we run this code the output is ['1', '2', '', '3', '', '', '4', '', '5', '6', '7', '', '8', '9', '', '10']
But i want to split the string with no blank like ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
Is there any function or method to remove blanks without using for loop like
for i in string:
if i == '':
string.remove(i)
Generate a list based on the output of split, and only include the elements which are not None
You can achieve this in multiple ways. The cleanest way here would be to use regex.
Regex:
import re
re.split('\++', inp)
#['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
List Comprehension:
inp = '1+2++3+++4++5+6+7++8+9++10'
[s for s in inp.split('+') if s]
#['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
Loop & Append:
result = []
for s in inp.split('+'):
if s:
result.append(s)
result
#['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
Simplest way:
customStr="1+2++3+++4++5+6+7++8+9++10"
list( filter( lambda x : x!="" ,customStr.split("+") ) )
n=10
fun=list(map(lambda x:[j for j in range(x)],n))
print(fun)
expected output is:
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
# using map (map works on an iterable)
n=10
fun=lambda x: (range(x)) # remember this is a function (so call it)
fun=list(map(str, fun(n)))
print(fun)
# directly
n=10
fun=list(map(str, (range(n))))
print(fun)
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
Python docs for reference:
map(): https://docs.python.org/3/library/functions.html#map
range(): https://docs.python.org/3/library/functions.html#func-range
lambda(): https://docs.python.org/3/library/ast.html?highlight=lambda#ast.Lambda
I need to read in a CSV file, from Excel, whose rows may be an arbitrary length.
The problem is the python retains these blank entries, but need to delete them for a future algorithm. Below is the output, I don't want the blank entries.
['5', '1', '5', '10', '4', '']
['3', '1', '5', '10', '2', '']
['6', '1', '5', '10', '5', '2']
['9', '10', '5', '10', '7', '']
['8', '5', '5', '10', '7', '']
['1', '1', '5', '10', '', '']
['2', '1', '5', '10', '1', '']
['7', '1', '5', '10', '6', '4']
['4', '1', '5', '10', '3', '1']
Here's a list comprehension integrated with the csv library:
import csv
with open('input.csv') as in_file:
reader = csv.reader(in_file)
result = [[item for item in row if item != ''] for row in reader]
print result
This is about as verbose a function as I could write to do what you want. There are certainly slicker ways.
def remove_blanks(a_list):
new_list = []
for item in a_list:
if item != "":
new_list.append(item)
return new_list
List comprehension version:
a = ['5', '1', '5', '10', '4', '']
[x for x in a if x != '']
Out[19]: ['5', '1', '5', '10', '4']
You may be better served by filtering at the csv read step instead.