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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
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("+") ) )
I have a list that contains a set of strings like this:
list = ['235,ACCESS,19841136,22564960,4291500,20,527434,566876','046,ALLOWED,24737321,27863065,1086500,3,14208500,14254500']
I'm trying to make the elements of the list a sublist but without splitting the string.
I tried new_list = list(map(list, list)). This is the result taking as reference the first element of the list:
print(new_list[0]):
[['2', '3', '5', ',', 'A', 'C', 'C', 'E', 'S',',','1', '9', '8', '4', '1', '1', '3', '6', ',', '2', '2', '5', '6', '4', '9', '6', '0', ',', '4', '2', '9', '1', '5', '0', '0', ',', '2', '0', ',', '5', '2', '7', '4', '3', '4', ',', '5', '6', '6', '8', '7', '6']]
I would like this output:
print(new_list[0]):
[[235,'ACCESS',19841136,22564960,4291500,20,527434,566876]]
Thanks in advance for your help!
You can try split() with delimiter , like this -
new_list = [i.split(',') for i in list]
print (new_list[0])
Output:
['235', 'ACCESS', '19841136', '22564960', '4291500', '20', '527434', '566876']
One thing is that here the numbers are also represented as string. If you want integers instead you can use isdigit() method like this -
new_list = [[int(e) if e.isdigit() else e for e in i.split(',') ]for i in list]
print(new_list[0])
Output:
[235, 'ACCESS', 19841136, 22564960, 4291500, 20, 527434, 566876]
Also, please try to avoid naming your list list
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).
I have a list of lists of strings:
listBefore = [['4', '5', '1', '1'],
['4', '6', '1', '1'],
['4', '7', '8', '1'],
['1', '2', '1', '1'],
['2', '3', '1', '1'],
['7', '8', '1', '1'],
['7', '9', '1', '1'],
['2', '4', '3', '1']]
and I would like to add a counter/index in the middle of each list so that the list looks like:
listAfter = [['4', '5', '1', '1', '1'],
['4', '6', '2', '1', '1'],
['4', '7', '3', '8', '1'],
['1', '2', '4', '1', '1'],
['2', '3', '5', '1', '1'],
['7', '8', '6', '1', '1'],
['7', '9', '7', '1', '1'],
['2', '4', '8', '3', '1']]
What would be the easiest way to do so? I could probably loop over the lists and add the index, but is there a cleaner way to do so?
Cheers,
Kate
Edit:
The code I wrote works for me:
item = 1
for list in listBefore:
list.insert(2,str(item))
item = item + 1
print listBefore
I was wondering if there is another way to do so more efficiently or in one step.
Iterate over the parent list with enumerate() to get counter (used as i in the below example) along with list element. In the sublist, insert the i at the middle of sublist using list.insert(index, value) method.
Note: The value of counter i.e. i will be of int type, so you have to explicitly type cast it to str as str(i)before inserting. Below is the sample code:
for i, sub_list in enumerate(my_list, 1): # Here my_list is the list mentioned in question as 'listBefore'
sub_list.insert(len(sub_list)/2, str(i))
# Value of 'my_list'
# [['4', '5', '1', '1', '1'],
# ['4', '6', '2', '1', '1'],
# ['4', '7', '3', '8', '1'],
# ['1', '2', '4', '1', '1'],
# ['2', '3', '5', '1', '1'],
# ['7', '8', '6', '1', '1'],
# ['7', '9', '7', '1', '1'],
# ['2', '4', '8', '3', '1']]
You can perform this with a list comprehension
listAfter = [listBefore[i][:len(listBefore[i])/2] + [str(i+1)] + listBefore[i][len(listBefore[i])/2:] for i in range(len(listBefore))]
You should learn about enumerate, it lets you iterate over the list with two iterators - one (str_list in this case) holds the current item in the list, and the other (i`) holds it's index in the list.
for i,str_list in enumerate(listBefore):
listBefore[i] = str_list[:len(str_list)//2] + [str(i+1)] + str_list[len(str_list)//2:]
>>> data = [['4', '5', '1', '1'],
['4', '6', '1', '1'],
['4', '7', '8', '1'],
['1', '2', '1', '1'],
['2', '3', '1', '1'],
['7', '8', '1', '1'],
['7', '9', '1', '1'],
['2', '4', '3', '1']]
>>> print [row[:len(row)//2] + [str(i)] + row[len(row)//2:]
for i, row in enumerate(data, start=1)]
[['4', '5', '1', '1', '1'],
['4', '6', '2', '1', '1'],
['4', '7', '3', '8', '1'],
['1', '2', '4', '1', '1'],
['2', '3', '5', '1', '1'],
['7', '8', '6', '1', '1'],
['7', '9', '7', '1', '1'],
['2', '4', '8', '3', '1']]