PythonTypeError: 'float' object is not iterable
I'm trying to do 3 slices from a list of vectors from a txt file, with 2 columns, using conditions to separate the RGB.
But when I run the program the following error appears: "'float' object is not iterable". Can anyone help me?
#Conditions
B = 0
G = 0
R = 0
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = list(vetor_x[i])
vetor_yB[B] = list(vetor_y[i])
B += 1
elif vetor_x[i] <= 600:
vetor_xG[G] = list(vetor_x[i])
vetor_yG[G] = list(vetor_y[i])
G += 1
elif vetor_x[i] <= 700:
vetor_xR[R] = list(vetor_x[i])
vetor_yR[R] = list(vetor_y[i])
R += 1
print('####### vetor_xB #######')
print(vetor_xB)
print('####### vetor_yB #######')
print(vetor_xB)
print('####### vetor_xG #######')
print(vetor_xG)
print('####### vetor_yG #######')
print(vetor_yG)
print('####### vetor_xR #######')
print(vetor_xR)
print('####### vetor_yR #######')
print(vetor_yR)
When I try to run it, this causes this error:
Traceback (most recent call last):
File "teste4.py", line 30, in <module>
vetor_xB[B] = list(vetor_x[i])
TypeError: 'float' object is not iterable
Please help me!
python python-3.x
asked Jan 2 at 6:14
FulanaFulana
324
|
show 3 more comments
I'm trying to do 3 slices from a list of vectors from a txt file, with 2 columns, using conditions to separate the RGB.
But when I run the program the following error appears: "'float' object is not iterable". Can anyone help me?
#Conditions
B = 0
G = 0
R = 0
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = list(vetor_x[i])
vetor_yB[B] = list(vetor_y[i])
B += 1
elif vetor_x[i] <= 600:
vetor_xG[G] = list(vetor_x[i])
vetor_yG[G] = list(vetor_y[i])
G += 1
elif vetor_x[i] <= 700:
vetor_xR[R] = list(vetor_x[i])
vetor_yR[R] = list(vetor_y[i])
R += 1
print('####### vetor_xB #######')
print(vetor_xB)
print('####### vetor_yB #######')
print(vetor_xB)
print('####### vetor_xG #######')
print(vetor_xG)
print('####### vetor_yG #######')
print(vetor_yG)
print('####### vetor_xR #######')
print(vetor_xR)
print('####### vetor_yR #######')
print(vetor_yR)
When I try to run it, this causes this error:
Traceback (most recent call last):
File "teste4.py", line 30, in <module>
vetor_xB[B] = list(vetor_x[i])
TypeError: 'float' object is not iterable
Please help me!
python python-3.x
asked Jan 2 at 6:14
FulanaFulana
324
what is yourvector_xB? Looks like it's a floating point.
– Kevin Fang
Jan 2 at 6:18
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
vetor_x.append(float(X))why into float?
– DirtyBit
Jan 2 at 6:26
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46
|
show 3 more comments
I'm trying to do 3 slices from a list of vectors from a txt file, with 2 columns, using conditions to separate the RGB.
But when I run the program the following error appears: "'float' object is not iterable". Can anyone help me?
#Conditions
B = 0
G = 0
R = 0
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = list(vetor_x[i])
vetor_yB[B] = list(vetor_y[i])
B += 1
elif vetor_x[i] <= 600:
vetor_xG[G] = list(vetor_x[i])
vetor_yG[G] = list(vetor_y[i])
G += 1
elif vetor_x[i] <= 700:
vetor_xR[R] = list(vetor_x[i])
vetor_yR[R] = list(vetor_y[i])
R += 1
print('####### vetor_xB #######')
print(vetor_xB)
print('####### vetor_yB #######')
print(vetor_xB)
print('####### vetor_xG #######')
print(vetor_xG)
print('####### vetor_yG #######')
print(vetor_yG)
print('####### vetor_xR #######')
print(vetor_xR)
print('####### vetor_yR #######')
print(vetor_yR)
When I try to run it, this causes this error:
Traceback (most recent call last):
File "teste4.py", line 30, in <module>
vetor_xB[B] = list(vetor_x[i])
TypeError: 'float' object is not iterable
Please help me!
python python-3.x
asked Jan 2 at 6:14
FulanaFulana
324
I'm trying to do 3 slices from a list of vectors from a txt file, with 2 columns, using conditions to separate the RGB.
But when I run the program the following error appears: "'float' object is not iterable". Can anyone help me?
#Conditions
B = 0
G = 0
R = 0
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = list(vetor_x[i])
vetor_yB[B] = list(vetor_y[i])
B += 1
elif vetor_x[i] <= 600:
vetor_xG[G] = list(vetor_x[i])
vetor_yG[G] = list(vetor_y[i])
G += 1
elif vetor_x[i] <= 700:
vetor_xR[R] = list(vetor_x[i])
vetor_yR[R] = list(vetor_y[i])
R += 1
print('####### vetor_xB #######')
print(vetor_xB)
print('####### vetor_yB #######')
print(vetor_xB)
print('####### vetor_xG #######')
print(vetor_xG)
print('####### vetor_yG #######')
print(vetor_yG)
print('####### vetor_xR #######')
print(vetor_xR)
print('####### vetor_yR #######')
print(vetor_yR)
When I try to run it, this causes this error:
Traceback (most recent call last):
File "teste4.py", line 30, in <module>
vetor_xB[B] = list(vetor_x[i])
TypeError: 'float' object is not iterable
Please help me!
python python-3.x
python python-3.x
asked Jan 2 at 6:14
FulanaFulana
324
asked Jan 2 at 6:14
FulanaFulana
324
edited Jan 2 at 21:19
Fulana
asked Jan 2 at 6:14
FulanaFulana
324
asked Jan 2 at 6:14
FulanaFulana
324
asked Jan 2 at 6:14
FulanaFulana
324
324
what is yourvector_xB? Looks like it's a floating point.
– Kevin Fang
Jan 2 at 6:18
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
vetor_x.append(float(X))why into float?
– DirtyBit
Jan 2 at 6:26
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46
|
show 3 more comments
what is yourvector_xB? Looks like it's a floating point.
– Kevin Fang
Jan 2 at 6:18
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
vetor_x.append(float(X))why into float?
– DirtyBit
Jan 2 at 6:26
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46
what is your
vector_xB? Looks like it's a floating point.– Kevin Fang
Jan 2 at 6:18
what is your
vector_xB? Looks like it's a floating point.– Kevin Fang
Jan 2 at 6:18
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
vetor_x.append(float(X)) why into float?– DirtyBit
Jan 2 at 6:26
vetor_x.append(float(X)) why into float?– DirtyBit
Jan 2 at 6:26
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46
|
show 3 more comments
1 Answer
1
active
oldest
votes
You cannot split an int or float type into a list.
You can convert them to str using str(vetor_x[i]) first if this is what you're going for.list() will attempt to split a string into each character.
For example, list('abc') will give you ['a','b','c'].
For int and float, this cannot be done because they are not iterables.
It does not look like you intend to split your vetor_x[i] into each character. It looks like you just want to store the values into vetor_xB[B], in which case, you should be creating an empty list with that many None or 0 variables and then your code to replace them with your
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = vetor_x[i]
vetor_yB[B] = vetor_y[i]
B += 1
.......
will work.
So you should actually create vetor_xB = [None]*500 to get vetor_xB = [None, None, x500...] before the above code will work
answered Jan 2 at 6:29
ycxycx
1,629417
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do avetor_x.sort()(optional if you want to sort or just store them as of the original order), Then do a directfor i in vetor_x:loop, check for the<=andappendthe floats tovetor_xBdirectly.
– ycx
Jan 2 at 7:03
add a comment |
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1 Answer
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1 Answer
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oldest
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active
oldest
votes
You cannot split an int or float type into a list.
You can convert them to str using str(vetor_x[i]) first if this is what you're going for.list() will attempt to split a string into each character.
For example, list('abc') will give you ['a','b','c'].
For int and float, this cannot be done because they are not iterables.
It does not look like you intend to split your vetor_x[i] into each character. It looks like you just want to store the values into vetor_xB[B], in which case, you should be creating an empty list with that many None or 0 variables and then your code to replace them with your
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = vetor_x[i]
vetor_yB[B] = vetor_y[i]
B += 1
.......
will work.
So you should actually create vetor_xB = [None]*500 to get vetor_xB = [None, None, x500...] before the above code will work
answered Jan 2 at 6:29
ycxycx
1,629417
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do avetor_x.sort()(optional if you want to sort or just store them as of the original order), Then do a directfor i in vetor_x:loop, check for the<=andappendthe floats tovetor_xBdirectly.
– ycx
Jan 2 at 7:03
add a comment |
You cannot split an int or float type into a list.
You can convert them to str using str(vetor_x[i]) first if this is what you're going for.list() will attempt to split a string into each character.
For example, list('abc') will give you ['a','b','c'].
For int and float, this cannot be done because they are not iterables.
It does not look like you intend to split your vetor_x[i] into each character. It looks like you just want to store the values into vetor_xB[B], in which case, you should be creating an empty list with that many None or 0 variables and then your code to replace them with your
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = vetor_x[i]
vetor_yB[B] = vetor_y[i]
B += 1
.......
will work.
So you should actually create vetor_xB = [None]*500 to get vetor_xB = [None, None, x500...] before the above code will work
answered Jan 2 at 6:29
ycxycx
1,629417
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do avetor_x.sort()(optional if you want to sort or just store them as of the original order), Then do a directfor i in vetor_x:loop, check for the<=andappendthe floats tovetor_xBdirectly.
– ycx
Jan 2 at 7:03
add a comment |
You cannot split an int or float type into a list.
You can convert them to str using str(vetor_x[i]) first if this is what you're going for.list() will attempt to split a string into each character.
For example, list('abc') will give you ['a','b','c'].
For int and float, this cannot be done because they are not iterables.
It does not look like you intend to split your vetor_x[i] into each character. It looks like you just want to store the values into vetor_xB[B], in which case, you should be creating an empty list with that many None or 0 variables and then your code to replace them with your
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = vetor_x[i]
vetor_yB[B] = vetor_y[i]
B += 1
.......
will work.
So you should actually create vetor_xB = [None]*500 to get vetor_xB = [None, None, x500...] before the above code will work
answered Jan 2 at 6:29
ycxycx
1,629417
You cannot split an int or float type into a list.
You can convert them to str using str(vetor_x[i]) first if this is what you're going for.list() will attempt to split a string into each character.
For example, list('abc') will give you ['a','b','c'].
For int and float, this cannot be done because they are not iterables.
It does not look like you intend to split your vetor_x[i] into each character. It looks like you just want to store the values into vetor_xB[B], in which case, you should be creating an empty list with that many None or 0 variables and then your code to replace them with your
for i in range(0,len(vetor_x)):
if vetor_x[i] <= 500:
vetor_xB[B] = vetor_x[i]
vetor_yB[B] = vetor_y[i]
B += 1
.......
will work.
So you should actually create vetor_xB = [None]*500 to get vetor_xB = [None, None, x500...] before the above code will work
answered Jan 2 at 6:29
ycxycx
1,629417
edited Jan 2 at 6:35
answered Jan 2 at 6:29
ycxycx
1,629417
answered Jan 2 at 6:29
ycxycx
1,629417
answered Jan 2 at 6:29
ycxycx
1,629417
1,629417
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do avetor_x.sort()(optional if you want to sort or just store them as of the original order), Then do a directfor i in vetor_x:loop, check for the<=andappendthe floats tovetor_xBdirectly.
– ycx
Jan 2 at 7:03
add a comment |
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do avetor_x.sort()(optional if you want to sort or just store them as of the original order), Then do a directfor i in vetor_x:loop, check for the<=andappendthe floats tovetor_xBdirectly.
– ycx
Jan 2 at 7:03
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
Hi @ycx, I want to split that array up there in pairs, by the x-coordinate (first column). Ex: 400 <= B> = 500, 500 <= B> = 600 and 600 <= B> = 700 and store them in new arrays. I do not know if I did not understand what you explained, but it still made a mistake.
– Fulana
Jan 2 at 6:55
@Fulana If that is the case, you can do a
vetor_x.sort() (optional if you want to sort or just store them as of the original order), Then do a direct for i in vetor_x: loop, check for the <= and append the floats to vetor_xB directly.– ycx
Jan 2 at 7:03
@Fulana If that is the case, you can do a
vetor_x.sort() (optional if you want to sort or just store them as of the original order), Then do a direct for i in vetor_x: loop, check for the <= and append the floats to vetor_xB directly.– ycx
Jan 2 at 7:03
add a comment |
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what is your
vector_xB? Looks like it's a floating point.– Kevin Fang
Jan 2 at 6:18
@KevinFang Is a list of values from the first column within the range> = 500
– Fulana
Jan 2 at 6:21
vetor_x.append(float(X))why into float?– DirtyBit
Jan 2 at 6:26
@user5173426 Because they are not integers
– Fulana
Jan 2 at 6:30
@Fulana I mean, what's it before this block of code, how do you define and initialize it.
– Kevin Fang
Jan 2 at 6:46