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python list of lists contain substring

I have the list_of_lists and I need to get the string that contains 'height' in the sublists and if there is no height at all I need to get 'nvt' for the whole sublist.
I have tried the following:
list_of_lists = [['width=9','length=3'],['width=6','length=4','height=4']]
_lists = []
for list in list_of_lists:
list1 = []
for st in list:
if ("height" ) in st:
list1.append(st)
else:
list1.append('nvt')
_lists.append(list1)
OUT = _lists
the result I need to have is :
_lists = ['nvt', 'height=4']
what I'm getting is:
_lists = [['nvt','nvt'],['nvt','nvt','height=4']]

This is a good case for implementing a for/else construct as follows:
list_of_lists = [['width=9','length=3'],['width=6','length=4','height=4']]
result = []
for e in list_of_lists:
for ss in e:
if ss.startswith('height'):
result.append(ss)
break
else:
result.append('nvt')
print(result)
Output:
['nvt', 'height=4']
Note:
This could probably be done with a list comprehension but I think this is more obvious and probably has no significant difference in terms of performance

This should work, you can assign height variable to first value in the sublist where s.startswith("height") is True, and if nothing matches this filter, you can assign height to 'nvt'.
_lists = []
for sublist in list_of_lists:
height = next(filter(lambda s: s.startswith("height"), sublist), 'nvt')
_lists.append(height)
And if you wish to be crazy, you can use list comprehension to reduce the code to the:
_lists = [next(filter(lambda s: s.startswith("height"), sublist), 'nvt') for sublist in list_of_lists]

Try this (Python 3.x):
import re
list_of_lists = [['width=9','length=3'],['width=6','length=4','height=4']]
_lists = []
r = re.compile("height=")
for li in list_of_lists:
match = list(filter(r.match, li))
if len(match) > 0:
_lists.extend(match)
else:
_lists.append('nvt')
OUT = _lists
print(OUT)

How to combine 3 lists to create a dictionary with one key and two values?

i would like to know how I could create a dictionary using the three lists. coun_keys to be a key and months_values and cases_values are to be the values.
I only found sources where I could use the zip() function to have a key: value, but how can I have key: value1, value2?
def main(csvfile,country ,type ):
with open(csvfile,"r") as file:
if type.lower() == "statistics ":
coun_keys = []
months_values = []
cases_values = []
listname =[]
coun_month={}
for line in file:
columns = (line.strip().split(","))
listname.append(columns)
listname.pop(0)
for line in listname:
date1 = line[3].split("/")
coun_keys.append(str(line[2]))
months_values.append(int(date1[1]))
cases_values.append(int(line[4]))

Do you mean, like:
list1 = [1, 2, 3]
list2 = 'abc'
list3 = [5, 6, 7]
print(dict(zip(list1,zip(list2, list3))))
#############################
For your code specifically, I would break up what you want to do into pieces. First define what you want to do with each line of your file:
def process_line(line):
line = line.strip().split(',')
date1 = line[3].split("/")
key = str(line[2])
month = int(date1[1])
case = int(line[4])
return key,(month,case)
Notice I group the values I want in a tuple, in particular, I want the process_line function to return my "key" and my "value" (a pair). Now open your file and process the lines:
f = open(csvfile)
next(f) #Skip the first line
result = dict(process_line(line) for line in f)
f.close()

This might help...
newDict = dict(zip(coun_keys, [months_values, cases_values]))

Assuming they're the same length, you can also do:
your_dict = {coun_keys[i] : (months_values[i], cases_values[i]) for i in range(len(coun_keys))}

For example,
lst1 = [a,b,c,d]
lst2 = [1,2,3,4]
lst3 = [5,6,7,8]
dict1 = dict(zip(lst1,zip(lst2,lst3)))

Python can't remove None value from list of lists

I'm trying to remove a None value from a csv file I have. I have converted blank values to None values in the first part of the below code but in the last part when I envoke filter It prints the column_list but the None values remain also. I need to remove them so I can work out max/min values of each which doesn't appear to work with them in the list?
with (inFile) as f:
_= next(f)
list_of_lists = [[float(i) if i.strip() != '' else None for i in line.split(',')[2:]]
for line in f]
inFile.close()
log_GDP_list = [item[0] for item in list_of_lists]
social_support_list = [item[1] for item in list_of_lists]
healthy_birth_list = [item[2] for item in list_of_lists]
freedom_choices_list = [item[3] for item in list_of_lists]
generosity_list = [item[4] for item in list_of_lists]
confidence_gov_list = [item[5] for item in list_of_lists]
column_list = []
column_list.append([log_GDP_list, social_support_list, healthy_birth_list, freedom_choices_list, generosity_list, confidence_gov_list])
res = list(filter(None, column_list))
print(res)
Also, when running the filter on just one of the row lists (such as log_GDP_list) it removes the None values but I still get an error saying I can't run max() or min() on floats (all values were converted from strings to floats in the first bit of the code).

You currently have something like this
l = [
float(i) if i.strip() != '' else None
for i in line.split(',')[2:]
]
what you want is this:
l = [
float(i)
for i in line.split(',')[2:]
if i.strip()
]
This way, when i.strip() evaluates to False, the item wont be added to the resulting list at all.

While loop within for loop for list of lists

I'm trying to create a big list that will contain lists of strings. I iterate over the input list of strings and create a temporary list.
Input:
['Mike','Angela','Bill','\n','Robert','Pam','\n',...]
My desired output:
[['Mike','Angela','Bill'],['Robert','Pam']...]
What i get:
[['Mike','Angela','Bill'],['Angela','Bill'],['Bill']...]
Code:
for i in range(0,len(temp)):
temporary = []
while(temp[i] != '\n' and i<len(temp)-1):
temporary.append(temp[i])
i+=1
bigList.append(temporary)

Use itertools.groupby
from itertools import groupby
names = ['Mike','Angela','Bill','\n','Robert','Pam']
[list(g) for k,g in groupby(names, lambda x:x=='\n') if not k]
#[['Mike', 'Angela', 'Bill'], ['Robert', 'Pam']]

Fixing your code, I'd recommend iterating over each element directly, appending to a nested list -
r = [[]]
for i in temp:
if i.strip():
r[-1].append(i)
else:
r.append([])
Note that if temp ends with a newline, r will have a trailing empty [] list. You can get rid of that though:
if not r[-1]:
del r[-1]
Another option would be using itertools.groupby, which the other answerer has already mentioned. Although, your method is more performant.

Your for loop was scanning over the temp array just fine, but the while loop on the inside was advancing that index. And then your while loop would reduce the index. This caused the repitition.
temp = ['mike','angela','bill','\n','robert','pam','\n','liz','anya','\n']
# !make sure to include this '\n' at the end of temp!
bigList = []
temporary = []
for i in range(0,len(temp)):
if(temp[i] != '\n'):
temporary.append(temp[i])
print(temporary)
else:
print(temporary)
bigList.append(temporary)
temporary = []

You could try:
a_list = ['Mike','Angela','Bill','\n','Robert','Pam','\n']
result = []
start = 0
end = 0
for indx, name in enumerate(a_list):
if name == '\n':
end = indx
sublist = a_list[start:end]
if sublist:
result.append(sublist)
start = indx + 1
>>> result
[['Mike', 'Angela', 'Bill'], ['Robert', 'Pam']]

Join sublist in a list in python when the sublists are still items in the big list

I have a list like this:
l = [[1,4], [3,6], [5,4]]
I want to join the inner list with ":". I want to have the result as:
l = ['1:4', '3:6', '5:4']
How can I achieve it with Python?

You can use list comprehension to achieve this:
[':'.join(map(str, x)) for x in l]

l = [[1,4], [3,6], [5,4]]
fl = []
for x in l:
fl.append(str(x[0])+':'+str(x[1]))
print(fl) # Final List, also you can do: l = fl
But, i think that you want to do dictionaries, if this is true, you have to do:
l = [[1,4], [3,6], [5,4]]
fd = {}
for x in l:
fd[x[0]] = x[1]
print(fd) # Final Dictionary
EXPLANATION:
l = [[1,4], [3,6], [5,4]] # Your list
fl = [] # New list (here will be the result)
for x in l: # For each item in the list l:
fl.append(str(x[0])+':'+str(x[1])) # Append the first part [0] of this item with ':' and the second part [1].
print(fl)
With Dictionaries:
l = [[1,4], [3,6], [5,4]] # Your list
fd = {} # New dictionary
for x in l: # For each item in the list l:
fd[x[0]] = x[1] # Make a key called with the first part of the item and a value with the second part.
print(fd) # Final Dictionary
Also you can make a dictionary easier:
l = [[1,4], [3,6], [5,4]]
l = dict(l) # Make a dictionary with every sublist of the list (it also works with tuples), the first part of the sublist is the key, and the second the value.

You can do:
final_list = []
for i in l:
a = str(i[0])+':'+str(i[1])
final_list.append(a)
print(final_list)