Here is another way to do this:
try:
id = testlist.index('1')
print testlist[id]
except ValueError:
print "Not Found"
If your list got large enough and you only expected to find the value in a sparse number of indices, consider that this code could execute much faster because you don't have to iterate every value in the list.
lookingFor = 1
i = 0
index = 0
try:
while i < len(testlist):
index = testlist.index(lookingFor,i)
i = index + 1
print index
except ValueError: #testlist.index() cannot find lookingFor
pass
If you expect to find the value a lot you should probably just append "index" to a list and print the list at the end to save time per iteration.
Use enumerate:
testlist = [1,2,3,5,3,1,2,1,6]
for position, item in enumerate(testlist):
if item == 1:
print position
I think that it might be useful to use the curselection() method from thte Tkinter library:
from Tkinter import *
listbox.curselection()
This method works on Tkinter listbox widgets, so you'll need to construct one of them instead of a list.
This will return a position like this:
('0',) (although later versions of Tkinter may return a list of ints instead)
Which is for the first position and the number will change according to the item position.
For more information, see this page: http://effbot.org/tkinterbook/listbox.htm
Greetings.
[x for x in range(len(testlist)) if testlist[x]==1]
Why complicate things?
testlist = [1,2,3,5,3,1,2,1,6]
for position, item in enumerate(testlist):
if item == 1:
print position
Here is another way to do this:
try:
id = testlist.index('1')
print testlist[id]
except ValueError:
print "Not Found"
Just to illustrate complete example along with the input_list
which has searies1
(example: input_list[0]) in which you want to do a lookup of series2
(example: input_list[1]) and get indexes of series2 if it exists in series1.
Note: Your certain condition
will go in lambda expression if conditions are simple
input_list = [[1,2,3,4,5,6,7],[1,3,7]]
series1 = input_list[0]
series2 = input_list[1]
idx_list = list(map(lambda item: series1.index(item) if item in series1 else None, series2))
print(idx_list)
output:
[0, 2, 6]
Try the below:
testlist = [1,2,3,5,3,1,2,1,6]
position=0
for i in testlist:
if i == 1:
print(position)
position=position+1
Just to illustrate complete example along with the input_list
which has searies1
(example: input_list[0]) in which you want to do a lookup of series2
(example: input_list[1]) and get indexes of series2 if it exists in series1.
Note: Your certain condition
will go in lambda expression if conditions are simple
input_list = [[1,2,3,4,5,6,7],[1,3,7]]
series1 = input_list[0]
series2 = input_list[1]
idx_list = list(map(lambda item: series1.index(item) if item in series1 else None, series2))
print(idx_list)
output:
[0, 2, 6]
What about the following?
print testlist.index(element)
If you are not sure whether the element to look for is actually in the list, you can add a preliminary check, like
if element in testlist:
print testlist.index(element)
or
print(testlist.index(element) if element in testlist else None)
or the "pythonic way", which I don't like so much because code is less clear, but sometimes is more efficient,
try:
print testlist.index(element)
except ValueError:
pass
Use enumerate:
testlist = [1,2,3,5,3,1,2,1,6]
for position, item in enumerate(testlist):
if item == 1:
print position
If your list got large enough and you only expected to find the value in a sparse number of indices, consider that this code could execute much faster because you don't have to iterate every value in the list.
lookingFor = 1
i = 0
index = 0
try:
while i < len(testlist):
index = testlist.index(lookingFor,i)
i = index + 1
print index
except ValueError: #testlist.index() cannot find lookingFor
pass
If you expect to find the value a lot you should probably just append "index" to a list and print the list at the end to save time per iteration.
Use enumerate:
testlist = [1,2,3,5,3,1,2,1,6]
for position, item in enumerate(testlist):
if item == 1:
print position
[x for x in range(len(testlist)) if testlist[x]==1]
testlist = [1,2,3,5,3,1,2,1,6]
for id, value in enumerate(testlist):
if id == 1:
print testlist[id]
I guess that it's exacly what you want. ;-) 'id' will be always the index of the values on the list.
testlist = [1,2,3,5,3,1,2,1,6]
for id, value in enumerate(testlist):
if id == 1:
print testlist[id]
I guess that it's exacly what you want. ;-) 'id' will be always the index of the values on the list.
for i in xrange(len(testlist)):
if testlist[i] == 1:
print i
xrange instead of range as requested (see comments).
Why complicate things?
testlist = [1,2,3,5,3,1,2,1,6]
for position, item in enumerate(testlist):
if item == 1:
print position
What about the following?
print testlist.index(element)
If you are not sure whether the element to look for is actually in the list, you can add a preliminary check, like
if element in testlist:
print testlist.index(element)
or
print(testlist.index(element) if element in testlist else None)
or the "pythonic way", which I don't like so much because code is less clear, but sometimes is more efficient,
try:
print testlist.index(element)
except ValueError:
pass
for i in xrange(len(testlist)):
if testlist[i] == 1:
print i
xrange instead of range as requested (see comments).
If your list got large enough and you only expected to find the value in a sparse number of indices, consider that this code could execute much faster because you don't have to iterate every value in the list.
lookingFor = 1
i = 0
index = 0
try:
while i < len(testlist):
index = testlist.index(lookingFor,i)
i = index + 1
print index
except ValueError: #testlist.index() cannot find lookingFor
pass
If you expect to find the value a lot you should probably just append "index" to a list and print the list at the end to save time per iteration.
I think that it might be useful to use the curselection() method from thte Tkinter library:
from Tkinter import *
listbox.curselection()
This method works on Tkinter listbox widgets, so you'll need to construct one of them instead of a list.
This will return a position like this:
('0',) (although later versions of Tkinter may return a list of ints instead)
Which is for the first position and the number will change according to the item position.
For more information, see this page: http://effbot.org/tkinterbook/listbox.htm
Greetings.
for i in xrange(len(testlist)):
if testlist[i] == 1:
print i
xrange instead of range as requested (see comments).
Source: Stackoverflow.com