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Having an issue appending to list inside function
How do I check if a list is empty?Finding the index of an item given a list containing it in PythonDifference between append vs. extend list methods in Pythonvar functionName = function() vs function functionName() How to append something to an array?Does Python have a ternary conditional operator?How to make a flat list out of list of lists?How to clone or copy a list?How do I list all files of a directory?Does Python have a string 'contains' substring method?
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I have the following table that has a unique ID, actual, and modeled output.
Pass_ID Actual Modeled
0 100 1 1
1 101 0 1
2 102 1 0
3 103 1 1
I wrote the following function so I can generate a list which eventually I'll turn into a dataframe so I can easily call back the rows for each Pass_ID to see if I can isolate why the model is not working for those rows. However, the list keeps coming back empty.
new_list = []
def isolate_diff(df):
if df['Modeled'].any() != df['Actual'].any():
new_list.appened(df['Pass_ID'])
else:
pass
return new_list
How can I fix this issue?
python list function append
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
add a comment |
I have the following table that has a unique ID, actual, and modeled output.
Pass_ID Actual Modeled
0 100 1 1
1 101 0 1
2 102 1 0
3 103 1 1
I wrote the following function so I can generate a list which eventually I'll turn into a dataframe so I can easily call back the rows for each Pass_ID to see if I can isolate why the model is not working for those rows. However, the list keeps coming back empty.
new_list = []
def isolate_diff(df):
if df['Modeled'].any() != df['Actual'].any():
new_list.appened(df['Pass_ID'])
else:
pass
return new_list
How can I fix this issue?
python list function append
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
Your code is incorrect. It should beappendinstead ofappenedin line 5.
– Yusufsn
Mar 8 at 2:04
You never call yourisolate_diff(df)function. Try addingnew_list = isolate_diff(df)to the bottom of your script
– Reedinationer
Mar 8 at 2:05
It is better to definenew_listinisolate_diffbecause the function will returnnew_list.
– Ellisein
Mar 8 at 2:08
add a comment |
I have the following table that has a unique ID, actual, and modeled output.
Pass_ID Actual Modeled
0 100 1 1
1 101 0 1
2 102 1 0
3 103 1 1
I wrote the following function so I can generate a list which eventually I'll turn into a dataframe so I can easily call back the rows for each Pass_ID to see if I can isolate why the model is not working for those rows. However, the list keeps coming back empty.
new_list = []
def isolate_diff(df):
if df['Modeled'].any() != df['Actual'].any():
new_list.appened(df['Pass_ID'])
else:
pass
return new_list
How can I fix this issue?
python list function append
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
I have the following table that has a unique ID, actual, and modeled output.
Pass_ID Actual Modeled
0 100 1 1
1 101 0 1
2 102 1 0
3 103 1 1
I wrote the following function so I can generate a list which eventually I'll turn into a dataframe so I can easily call back the rows for each Pass_ID to see if I can isolate why the model is not working for those rows. However, the list keeps coming back empty.
new_list = []
def isolate_diff(df):
if df['Modeled'].any() != df['Actual'].any():
new_list.appened(df['Pass_ID'])
else:
pass
return new_list
How can I fix this issue?
python list function append
python list function append
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
edited Mar 8 at 2:27
Pikachu the Purple Wizard
2,03061429
2,03061429
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
asked Mar 8 at 2:01
Ryan SullivanRyan Sullivan
82
82
Your code is incorrect. It should beappendinstead ofappenedin line 5.
– Yusufsn
Mar 8 at 2:04
You never call yourisolate_diff(df)function. Try addingnew_list = isolate_diff(df)to the bottom of your script
– Reedinationer
Mar 8 at 2:05
It is better to definenew_listinisolate_diffbecause the function will returnnew_list.
– Ellisein
Mar 8 at 2:08
add a comment |
Your code is incorrect. It should beappendinstead ofappenedin line 5.
– Yusufsn
Mar 8 at 2:04
You never call yourisolate_diff(df)function. Try addingnew_list = isolate_diff(df)to the bottom of your script
– Reedinationer
Mar 8 at 2:05
It is better to definenew_listinisolate_diffbecause the function will returnnew_list.
– Ellisein
Mar 8 at 2:08
Your code is incorrect. It should be
append instead of appened in line 5.– Yusufsn
Mar 8 at 2:04
Your code is incorrect. It should be
append instead of appened in line 5.– Yusufsn
Mar 8 at 2:04
You never call your
isolate_diff(df) function. Try adding new_list = isolate_diff(df) to the bottom of your script– Reedinationer
Mar 8 at 2:05
You never call your
isolate_diff(df) function. Try adding new_list = isolate_diff(df) to the bottom of your script– Reedinationer
Mar 8 at 2:05
It is better to define
new_list in isolate_diff because the function will return new_list.– Ellisein
Mar 8 at 2:08
It is better to define
new_list in isolate_diff because the function will return new_list.– Ellisein
Mar 8 at 2:08
add a comment |
3 Answers
3
active
oldest
votes
I agree with @Prune, but to solve it, i think you need a whole different solution:
print(df[df['Modeled']!=df['Actual']])
Output:
Pass_ID Actual Modeled
1 101 0 1
2 102 1 0
If you want a list of Pass_ID:
print(df.loc[df['Modeled']!=df['Actual'],'Pass_ID'].tolist())
Output:
[101, 102]
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
add a comment |
You wrote a global condition, not a vectorized condition:
if df['Modeled'].any() != df['Actual'].any():
Each term is True if any element is 1. This evaluates quickly to
if True != True
which gets you an empty new_list.
Instead, write an expression grabbing df['Pass_ID'], filtered where df['Modeled'] != df['Actual']. Can you take it from there?
answered Mar 8 at 2:08
PrunePrune
45.8k143559
add a comment |
Change new_list.appened(df['Pass_ID']) to new_list.append(df['Pass_ID'])
Edit: Prune and U9 detected a few other errors. This was the first incorrect thing I saw. Please defer to their answers for more detailed revisions
answered Mar 8 at 2:09
alec935alec935
945624
That's not all tho...
– U9-Forward
Mar 8 at 2:15
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I agree with @Prune, but to solve it, i think you need a whole different solution:
print(df[df['Modeled']!=df['Actual']])
Output:
Pass_ID Actual Modeled
1 101 0 1
2 102 1 0
If you want a list of Pass_ID:
print(df.loc[df['Modeled']!=df['Actual'],'Pass_ID'].tolist())
Output:
[101, 102]
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
add a comment |
I agree with @Prune, but to solve it, i think you need a whole different solution:
print(df[df['Modeled']!=df['Actual']])
Output:
Pass_ID Actual Modeled
1 101 0 1
2 102 1 0
If you want a list of Pass_ID:
print(df.loc[df['Modeled']!=df['Actual'],'Pass_ID'].tolist())
Output:
[101, 102]
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
add a comment |
I agree with @Prune, but to solve it, i think you need a whole different solution:
print(df[df['Modeled']!=df['Actual']])
Output:
Pass_ID Actual Modeled
1 101 0 1
2 102 1 0
If you want a list of Pass_ID:
print(df.loc[df['Modeled']!=df['Actual'],'Pass_ID'].tolist())
Output:
[101, 102]
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
I agree with @Prune, but to solve it, i think you need a whole different solution:
print(df[df['Modeled']!=df['Actual']])
Output:
Pass_ID Actual Modeled
1 101 0 1
2 102 1 0
If you want a list of Pass_ID:
print(df.loc[df['Modeled']!=df['Actual'],'Pass_ID'].tolist())
Output:
[101, 102]
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
answered Mar 8 at 2:11
U9-ForwardU9-Forward
17.9k51743
17.9k51743
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
add a comment |
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
1
1
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
wow. that's awesome. thanks so much. I think I over though it. really appreciate everyone's help
– Ryan Sullivan
Mar 8 at 4:35
add a comment |
You wrote a global condition, not a vectorized condition:
if df['Modeled'].any() != df['Actual'].any():
Each term is True if any element is 1. This evaluates quickly to
if True != True
which gets you an empty new_list.
Instead, write an expression grabbing df['Pass_ID'], filtered where df['Modeled'] != df['Actual']. Can you take it from there?
answered Mar 8 at 2:08
PrunePrune
45.8k143559
add a comment |
You wrote a global condition, not a vectorized condition:
if df['Modeled'].any() != df['Actual'].any():
Each term is True if any element is 1. This evaluates quickly to
if True != True
which gets you an empty new_list.
Instead, write an expression grabbing df['Pass_ID'], filtered where df['Modeled'] != df['Actual']. Can you take it from there?
answered Mar 8 at 2:08
PrunePrune
45.8k143559
add a comment |
You wrote a global condition, not a vectorized condition:
if df['Modeled'].any() != df['Actual'].any():
Each term is True if any element is 1. This evaluates quickly to
if True != True
which gets you an empty new_list.
Instead, write an expression grabbing df['Pass_ID'], filtered where df['Modeled'] != df['Actual']. Can you take it from there?
answered Mar 8 at 2:08
PrunePrune
45.8k143559
You wrote a global condition, not a vectorized condition:
if df['Modeled'].any() != df['Actual'].any():
Each term is True if any element is 1. This evaluates quickly to
if True != True
which gets you an empty new_list.
Instead, write an expression grabbing df['Pass_ID'], filtered where df['Modeled'] != df['Actual']. Can you take it from there?
answered Mar 8 at 2:08
PrunePrune
45.8k143559
answered Mar 8 at 2:08
PrunePrune
45.8k143559
answered Mar 8 at 2:08
PrunePrune
45.8k143559
answered Mar 8 at 2:08
PrunePrune
45.8k143559
45.8k143559
add a comment |
add a comment |
Change new_list.appened(df['Pass_ID']) to new_list.append(df['Pass_ID'])
Edit: Prune and U9 detected a few other errors. This was the first incorrect thing I saw. Please defer to their answers for more detailed revisions
answered Mar 8 at 2:09
alec935alec935
945624
That's not all tho...
– U9-Forward
Mar 8 at 2:15
add a comment |
Change new_list.appened(df['Pass_ID']) to new_list.append(df['Pass_ID'])
Edit: Prune and U9 detected a few other errors. This was the first incorrect thing I saw. Please defer to their answers for more detailed revisions
answered Mar 8 at 2:09
alec935alec935
945624
That's not all tho...
– U9-Forward
Mar 8 at 2:15
add a comment |
Change new_list.appened(df['Pass_ID']) to new_list.append(df['Pass_ID'])
Edit: Prune and U9 detected a few other errors. This was the first incorrect thing I saw. Please defer to their answers for more detailed revisions
answered Mar 8 at 2:09
alec935alec935
945624
Change new_list.appened(df['Pass_ID']) to new_list.append(df['Pass_ID'])
Edit: Prune and U9 detected a few other errors. This was the first incorrect thing I saw. Please defer to their answers for more detailed revisions
answered Mar 8 at 2:09
alec935alec935
945624
edited Mar 8 at 2:24
answered Mar 8 at 2:09
alec935alec935
945624
answered Mar 8 at 2:09
alec935alec935
945624
answered Mar 8 at 2:09
alec935alec935
945624
945624
That's not all tho...
– U9-Forward
Mar 8 at 2:15
add a comment |
That's not all tho...
– U9-Forward
Mar 8 at 2:15
That's not all tho...
– U9-Forward
Mar 8 at 2:15
That's not all tho...
– U9-Forward
Mar 8 at 2:15
add a comment |
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Your code is incorrect. It should be
appendinstead ofappenedin line 5.– Yusufsn
Mar 8 at 2:04
You never call your
isolate_diff(df)function. Try addingnew_list = isolate_diff(df)to the bottom of your script– Reedinationer
Mar 8 at 2:05
It is better to define
new_listinisolate_diffbecause the function will returnnew_list.– Ellisein
Mar 8 at 2:08