The "data" is a text record with several numbers in two columns.
x = np.linspace(1735.0,1775.0,100)column1 = (data<0,0:-1>+data<0,1:>)/2.0column2 = data<1,1:>x_column1 = np.zeros(x.size+2)x_column1<1:-1> = xx_column1<0> = x<0>+x<0>-x<1>x_column1<-1> = x<-1>+x<-1>-x<-2>experiment = np.zeros_like(x)for i in range(np.size(x_edges)-2): indexes = np.flatnonzero(np.logical_and((column1>=x_column1),(column1
python numpy indexing error-handling index-error
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edited Feb 8 "19 at 22:17
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asked jan 5 "17 in ~ 18:37
Seoyeon HongSeoyeon Hong
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In numpy, index and dimension number is numbered starts through 0. For this reason axis 0 means the 1st dimension. Likewise in numpy a measurement can have actually length (size) 0. The simplest instance is:
In <435>: x = np.zeros((0,), int)In <436>: xOut<436>: array(<>, dtype=int32)In <437>: x<0>...IndexError: index 0 is the end of bounds for axis 0 with dimension 0I additionally get it if x = np.zeros((0,5), int), a 2d variety with 0 rows, and also 5 columns.
You are watching: Indexerror: index 0 is out of bounds for axis 0 with size 0
So someplace in your password you room creating selection with a size 0 first axis.
When asking about errors, it is meant that girlfriend tell united state where the error occurs.
Also when debugging troubles like this, the very first thing you have to do is print the shape (and maybe the dtype) the the doubt variables.
Applied to pandasThe same error can take place for those using pandas, once sending a series or DataFrame come a numpy.array, similar to the following:
Resolving the error:Use a try-except blockVerify the dimension of the variety is no 0if x.size != 0:
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edited jan 12 at 0:41
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answered january 5 "17 in ~ 19:02
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Essentially it way you don"t have actually the index you are trying come reference. Because that example:
df = pd.DataFrame()df<"this">=np.nandf<"my">=np.nandf<"data">=np.nandf<"data"><0>=5 #I haven"t yet assigned how long df must be!print(df)will give me the error you are referring to, since I haven"t said Pandas just how long my dataframe is. Whereas if I carry out the precise same code however I perform assign an index length, ns don"t get an error:
df = pd.DataFrame(index=<0,1,2,3,4>)df<"this">=np.nandf<"is">=np.nandf<"my">=np.nandf<"data">=np.nandf<"data"><0>=5 #since I"ve correctly labelled mine index, ns don"t run right into this problem!print(df)Hope that answers her question!
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reply Apr 27 "18 in ~ 21:29
Daniel AbudDaniel Abud
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This is an IndexError in python, which way that we"re trying to accessibility an table of contents which isn"t over there in the tensor. Listed below is a very an easy example to recognize this error.
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# produce an empty array of dimension `0`In <14>: arr = np.array(<>, dtype=np.int64) # inspect its form In <15>: arr.shape Out<15>: (0,)with this range arr in place, if us now shot to assign any type of value to some index, for example to the index 0 together in the instance below
In <16>: arr<0> = 23 Then, we will acquire an IndexError, as below:
IndexError Traceback (most recent call last) in ----> 1 arr<0> = 23IndexError: table of contents 0 is out of bounds because that axis 0 with dimension 0The reason is the we are trying to accessibility an table of contents (here at 0th position), i m sorry is not there (i.e. That doesn"t exist due to the fact that we have an array of size 0).
In <19>: arr.size * arr.itemsize Out<19>: 0So, in essence, such an array is useless and cannot be supplied for save on computer anything. Thus, in her code, you"ve to monitor the traceback and look because that the place where you"re producing an array/tensor of size 0 and also fix that.