How to count number of 0 in a list of lists?
Question:
I want calculate the number of time a 0 appears in a "list of lists" web scraped, it’s like my list of lists is not a list because when I print, it’s done that :
[3]
[1]
[3]
[3]
[1]
[1]
[0]
[0]
[1]
[1]
if it’s really is a list of list it’s should tell me this :
[[3],[1],[3],[3],[1],[1],[0],[0],[1],[1]]
So, list or not list I cant count the number of time the 0 is here, how can I do for count the 0 ?
In this exemple, there is two 0, so the result is = 2
There is my code, maybe you can tell me how I can, I tried count(0) and other method, but it’s dont work
from bs4 import BeautifulSoup
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
print(infoss) is :
3
1
3
3
1
1
0
0
1
1
and i need a code which tell me the number of 0 in this list, so here it’s 2
Answers:
You are already iterating through the list of lists and printing out the values within it.
Can you think of a way to check the value instead of printing it and keep track of the number of times the value is equal to 0?
EDIT: Are you saying that this code doesn’t work?
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
coun = 0
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
if infoss == 0 :
coun = coun + 1
print(coun)
i tried this :
from bs4 import BeautifulSoup
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
coun = 0
if infoss < 1 :
coun = coun + 1
print(coun)
the result is this :
0
0
0
0
0
0
1
1
0
0
yes it’s tell me 1 and 1, but i dont want that form of answer, i just want it’s tell me result = 2, without all the rest
i tried to just have the number of item with a len, but it’s seems to me, it’s a bunch of list, but they are not in big list as they should to be. I cant track nothing, they seems not linked betwwen them in a big list.
I tried a dico method, and doesnt work too
You could try creating new list to store value of infoss then count element infoss that has 0.
lst_infoss = []
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
# append infoss to new list (lst_infoss)
lst_infoss.append(infoss)
print(lst_infoss)
zero_infoss = lst_infoss.count(0)
print(zero_infoss)
You were close to your goal and based on question title and your second example you could also use a list comprehension iterating your list and check the len() of result:
ls = [[3],[1],[3],[3],[1],[1],[0],[0],[1],[1]]
len([l for l in ls if l[0] == 0])
#output -> 2
Note: Avoid using python reserved terms (keywords), this could have unwanted effects on the results of your code.
I want calculate the number of time a 0 appears in a "list of lists" web scraped, it’s like my list of lists is not a list because when I print, it’s done that :
[3]
[1]
[3]
[3]
[1]
[1]
[0]
[0]
[1]
[1]
if it’s really is a list of list it’s should tell me this :
[[3],[1],[3],[3],[1],[1],[0],[0],[1],[1]]
So, list or not list I cant count the number of time the 0 is here, how can I do for count the 0 ?
In this exemple, there is two 0, so the result is = 2
There is my code, maybe you can tell me how I can, I tried count(0) and other method, but it’s dont work
from bs4 import BeautifulSoup
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
print(infoss) is :
3
1
3
3
1
1
0
0
1
1
and i need a code which tell me the number of 0 in this list, so here it’s 2
You are already iterating through the list of lists and printing out the values within it.
Can you think of a way to check the value instead of printing it and keep track of the number of times the value is equal to 0?
EDIT: Are you saying that this code doesn’t work?
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
coun = 0
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
if infoss == 0 :
coun = coun + 1
print(coun)
i tried this :
from bs4 import BeautifulSoup
import requests
from csv import writer
url= "https://blablablabla.htm"
page = requests.get(url)
soup = BeautifulSoup(page.content, 'html.parser')
lists = soup.find_all('tr', class_="mat1")
with open('file.csv', 'w', encoding='utf8', newline='') as f:
thewriter = writer(f)
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
coun = 0
if infoss < 1 :
coun = coun + 1
print(coun)
the result is this :
0
0
0
0
0
0
1
1
0
0
yes it’s tell me 1 and 1, but i dont want that form of answer, i just want it’s tell me result = 2, without all the rest
i tried to just have the number of item with a len, but it’s seems to me, it’s a bunch of list, but they are not in big list as they should to be. I cant track nothing, they seems not linked betwwen them in a big list.
I tried a dico method, and doesnt work too
You could try creating new list to store value of infoss then count element infoss that has 0.
lst_infoss = []
for list in lists:
but = list.find('td', class_="tj1").text.replace('xa0', '').replace(" ", "1")
info = but[3:]
infos = len(info)
infoss = int(infos)
print(infoss)
# append infoss to new list (lst_infoss)
lst_infoss.append(infoss)
print(lst_infoss)
zero_infoss = lst_infoss.count(0)
print(zero_infoss)
You were close to your goal and based on question title and your second example you could also use a list comprehension iterating your list and check the len() of result:
ls = [[3],[1],[3],[3],[1],[1],[0],[0],[1],[1]]
len([l for l in ls if l[0] == 0])
#output -> 2
Note: Avoid using python reserved terms (keywords), this could have unwanted effects on the results of your code.