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S_A_V [24]
2 years ago
7

What kind of internal factors influence the process of business enterprise

Computers and Technology
1 answer:
fgiga [73]2 years ago
4 0

Answer:

The three main internal factors are:

Human resources

Finance

Current or modern technology

Explanation:

Financial resources like funding, investment opportunities, and sources of income.

Physical resources like the company's location, equipment, and facilities.

Human resources like employees, target audiences, and volunteers.

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Answer:

Code

import csv

def open_file():

try:

fname=input("Enter the filename to open (xxx.yyy): ")

fp=open(fname,encoding='utf-8')

return fp

except FileNotFoundError:

print("File not found! Please Enter correct filename with extension")

return open_file()

def read_file(fp):

fields = []

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csvreader = list(csv.reader(fp))

# skiped 1 row because it contains field name

fields = csvreader[0]

D1={}

D2={}

D3={}

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else:

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na_sales = float(line[5])

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D1[name]=[]

D1[name].append((name, platform, year, genre, publisher,global_sales))

if genre not in D2:

D2[genre]=[]

D2[genre].append((genre, year, na_sales, eur_sales,jpn_sales, other_sales, global_sales))

if publisher not in D3:

D3[publisher]=[]

D3[publisher].append((publisher, name, year, na_sales,eur_sales, jpn_sales, other_sales, global_sales))

# sorting

temp={}

# sorting keys

for Keys in sorted (D1) :

ls=D1[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D1=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D2) :

ls=D2[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D2=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D3) :

ls=D3[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

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def main():

fp=open_file()

D1,D2,D3=read_file(fp)

# displaying the row_count in dictionaries

cnt=0

for d in D1:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D2:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D3:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

 

 

if __name__=="__main__":

main()

Explanation:

see code and see output attached

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