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Python Unit testing - IndexError: list index out of range


By : Marcus
Date : October 16 2020, 08:10 PM
Does that help your values variable has only one member which is a list of 5 int! so values[i] can not go further than values[0]
you can change values variable to list of 5 int like this:
code :
values = [1, 33, 1, 1, 1]  
values = (1, 33, 1, 1, 1) 
json = {columns[i]: values[0][i] for i in range(len(columns))}  


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IndexError: list index out of range [Python]


By : Sonai
Date : March 29 2020, 07:55 AM
To fix this issue There is something wrong with that piece of code, the "if name == 'main':" statement, means that the code inside the "if" only work when the python code is executed and not when it is used as a module. But inside the "if", a recursive call: hash_file(sys.argv[1]) is used, it means that the code needs an argument, but it will start a infinite recursive loop.
I think that the code:
code :
if __name__ == '__main__':
    hash_file(sys.argv[1]);
import sys, hashlib, os

def hash_file(filename):    #Calculate MD5 and SHA1 hash values of a given file

# Create hash objects for MD5 and SHA1.
md5_hash = hashlib.md5()
sha1_hash = hashlib.sha1()
filename = r"C:/this.png"

# Read the given file by 2K blocks. Feed blocks
# into into the hash objects by "update(data)" method.
fp = open(filename,'rb')
while 1:
    data = fp.read(2048)
    if not data:
        break
    else:
        md5_hash.update(data)
        sha1_hash.update(data)
fp.close()
print "The MD5  hash of your file is" 
print filename,":", md5_hash.hexdigest();
print "The SHA1 hash of your file is" 
print filename,":", sha1_hash.hexdigest();

# other code here

if __name__ == '__main__':
 #hash_file(sys.argv[1]);
 hash_file(woord)

python 3 IndexError: list index out of range


By : user96966
Date : March 29 2020, 07:55 AM
Hope that helps When you call a_list.append(row[0:8]) you're appending an array using only indexes 0, 1, 2, 3, 4, 5, 6, and 7 from row. This means that when you later iterate a_list, the x variable only has indexes up to 7, and you're trying to access 8.
Quick example:
code :
>>> row = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> x = row[:8]
>>> x[8]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range
>>> x[7]
7
>>>

Index out of range error in Python (IndexError: list index out of range)


By : ChipDesigner
Date : March 29 2020, 07:55 AM
it should still fix some issue I can't seem to find the issue here that I am getting out of range issue: , This should help:
code :
layerZ = [layer_1,layer_2,layer_3,layer_4,layer_5,layer_6,layer_7,layer_8,layer_9,layer_10,layer_11,layer_12,layer_13]

layerZ_total = []
layerZ_sp = []
layerZ_nonSp = []


for x in range(0, 12):
    layerZ_total.append(np.size(layerZ[x]))
    layerZ_sp.append(np.count_nonzero(layerZ[x]==0))
    layerZ_nonSp.append(np.count_nonzero(layerZ[x]))

    #Printing the results on scree to trace
    print "Layer:",x+1,"Threshhold:",repr(ths),"Total Parameters: ",layerZ_total[x],"# Sp: ",layerZ_sp[x],"# Remained : ",layerZ_nonSp[x],"Sp %: ",float(layerZ_sp[x])/layerZ_total[x]

IndexError: List index out of range - Python CSV


By : Bruce Galvez
Date : March 29 2020, 07:55 AM
I wish this help you As Ken White pointed out in the comments above. The error is caused by you trying to access an index that is outside the bounds of the list.
What is going on is that there is a blank row in your CSV file that python cannot process because you are calling index 0 even though it does not exist therefore python throws an exception.
code :
if(len(row) < 1):
   continue
ks={{tweet,tweetyouwant,tweet},{list,two,if,present}}
ks={tweet,tweetyouwant,tweet}
def tweet6(self):
    with codecs.open('HELLOTWITTER.csv', 'r', encoding='utf-8', errors='ignore') as f:
        reader = csv.reader(f)
        d = {}
        for i, row in enumerate(reader):

            #Verify row is within range
            if(len(row) < 1):
                continue

            #Get the rows values
            d[row[0]]=row[1:]

            #If past row 10 then break
            if (i>=10):
                break  

    #ks=list(d) #Not needed D is already a list
    return (d[5]) #return the row of the 6th tweet

IndexError: list index out of range when in python, even when the range exists


By : Akbar Pardayev
Date : March 29 2020, 07:55 AM
it helps some times The error is because you are not adding assessment value to sumc list for the last student after the loop ends. So, for n unique student id, the list length is only n-1. After for loop, add sumc.append(count). See below.
code :
assessment = assessment.values

count=0
stucount=28400
sumc=[]
i=0
for stu in assessment[:,2:3]:
    if(stucount==stu):
        count = count + assessment[i,5]
        i=i+1
    else:
        sumc.append(count)
        count = 0
        count = count + assessment[i,5]
        i=i+1
    stucount=stu

sumc.append(count)
print(sumc)

stucount=28400
i=0
a=[]
for stu in assessment[:,2:3]:
    if(stucount==stu):
        a.append(sumc[i])
        stucount = stu
    else:
        a.append(sumc[i])
        stucount = stu
        i=i+1

print(a)
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