How to append into a list if it doesn't find anything that matches the list?
I want to make a print when my code doesn't find anything throughout the whole list.
This is my code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
condition = True
while condition:
try:
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
condition = False
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_lis
condition = False
else:
old_list.append(new_list)
condition = False
break
except Exception as err:
print(err)
What I want do is: check every first word of each % and then I make a compare from old_list to check if there is the first word inside the old_list - if there is one then we replay it the old_list value with the new one.
elif if the word is longer than then the new_list word then we dont do anything.
How can I make so that if it doesn't find if a[0] in items: at the whole list then we add the new_list into old_list bascially?
Meaning it needs to first search for all words if it contains inside the old_list and if it is not then we append.
New edit:
Code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
break
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_list
break
else:
old_list.append(new_list)
break
python for-loop while-loop
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
add a comment |
I want to make a print when my code doesn't find anything throughout the whole list.
This is my code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
condition = True
while condition:
try:
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
condition = False
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_lis
condition = False
else:
old_list.append(new_list)
condition = False
break
except Exception as err:
print(err)
What I want do is: check every first word of each % and then I make a compare from old_list to check if there is the first word inside the old_list - if there is one then we replay it the old_list value with the new one.
elif if the word is longer than then the new_list word then we dont do anything.
How can I make so that if it doesn't find if a[0] in items: at the whole list then we add the new_list into old_list bascially?
Meaning it needs to first search for all words if it contains inside the old_list and if it is not then we append.
New edit:
Code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
break
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_list
break
else:
old_list.append(new_list)
break
python for-loop while-loop
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25
add a comment |
I want to make a print when my code doesn't find anything throughout the whole list.
This is my code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
condition = True
while condition:
try:
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
condition = False
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_lis
condition = False
else:
old_list.append(new_list)
condition = False
break
except Exception as err:
print(err)
What I want do is: check every first word of each % and then I make a compare from old_list to check if there is the first word inside the old_list - if there is one then we replay it the old_list value with the new one.
elif if the word is longer than then the new_list word then we dont do anything.
How can I make so that if it doesn't find if a[0] in items: at the whole list then we add the new_list into old_list bascially?
Meaning it needs to first search for all words if it contains inside the old_list and if it is not then we append.
New edit:
Code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
break
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_list
break
else:
old_list.append(new_list)
break
python for-loop while-loop
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
I want to make a print when my code doesn't find anything throughout the whole list.
This is my code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
condition = True
while condition:
try:
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
condition = False
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_lis
condition = False
else:
old_list.append(new_list)
condition = False
break
except Exception as err:
print(err)
What I want do is: check every first word of each % and then I make a compare from old_list to check if there is the first word inside the old_list - if there is one then we replay it the old_list value with the new one.
elif if the word is longer than then the new_list word then we dont do anything.
How can I make so that if it doesn't find if a[0] in items: at the whole list then we add the new_list into old_list bascially?
Meaning it needs to first search for all words if it contains inside the old_list and if it is not then we append.
New edit:
Code:
new_list = ['There%is', 'None&of', 'Same&here']
a = new_list.split("%")
old_list = ['Stack%Hello', 'Over&You', 'flow&There']
for i, items in zip(range(len(old_list)), old_list):
old_list_value = old_list[i].split("%")
if a[0] in items:
if len(old_list_value[1]) < len(a[1]):
old_list[i] = new_list
break
elif len(old_list_value[1]) > len(a[1]):
old_list[i] = new_list
break
else:
old_list.append(new_list)
break
python for-loop while-loop
python for-loop while-loop
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
edited Nov 21 '18 at 23:23
Hellosiroverthere
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
asked Nov 21 '18 at 23:12
HellosiroverthereHellosiroverthere
10818
10818
You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25
add a comment |
You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25
You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25
You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25
add a comment |
1 Answer
1
active
oldest
votes
You might want to check out the for-else:
for item in items:
if you_found_something:
found_item = item
break
else:
print('No item found.')
do_your_append_here
Basically the for-else answers the question if the for loop terminated without a break. Above we break when we find something therefore the else won't be executed but as far as your for loop terminates without a break statement, then the code in else will be executed.
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
active
oldest
votes
You might want to check out the for-else:
for item in items:
if you_found_something:
found_item = item
break
else:
print('No item found.')
do_your_append_here
Basically the for-else answers the question if the for loop terminated without a break. Above we break when we find something therefore the else won't be executed but as far as your for loop terminates without a break statement, then the code in else will be executed.
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
add a comment |
You might want to check out the for-else:
for item in items:
if you_found_something:
found_item = item
break
else:
print('No item found.')
do_your_append_here
Basically the for-else answers the question if the for loop terminated without a break. Above we break when we find something therefore the else won't be executed but as far as your for loop terminates without a break statement, then the code in else will be executed.
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
add a comment |
You might want to check out the for-else:
for item in items:
if you_found_something:
found_item = item
break
else:
print('No item found.')
do_your_append_here
Basically the for-else answers the question if the for loop terminated without a break. Above we break when we find something therefore the else won't be executed but as far as your for loop terminates without a break statement, then the code in else will be executed.
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
You might want to check out the for-else:
for item in items:
if you_found_something:
found_item = item
break
else:
print('No item found.')
do_your_append_here
Basically the for-else answers the question if the for loop terminated without a break. Above we break when we find something therefore the else won't be executed but as far as your for loop terminates without a break statement, then the code in else will be executed.
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
edited Nov 21 '18 at 23:19
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
answered Nov 21 '18 at 23:16
RafaelRafael
2,88932130
2,88932130
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
add a comment |
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
In that case I dont need to actually do a while loop and stuff that I did?
– Hellosiroverthere
Nov 21 '18 at 23:18
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
That's correct. You might want to restructure your solution with this new syntax in mind, which should solve your problem.
– Rafael
Nov 21 '18 at 23:19
add a comment |
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You don't need a break statement in the else.
– Rafael
Nov 21 '18 at 23:25