Question

Assume that an array A is given to you in a txt file. You should read it. If you think this is a time-dependent data, print the number of local minimums in the array to a txt file. Local minimum means the element which is less than the previous element and the following element. Please do not use any additional library. For example, if A=[3, 2, 9, 8, 7, 8, 6], the code should print 2 and 7. (Local minimums are 2 and 7).

write in python .

Answer 1

#source code withiut using file:

data=[3,2,9,8,7,8,6]

local_minimum_count=0

for i in range(1,len(data)-1):

if(data[i-1]>data[i] and data[i+1]>data[i]):

local_minimum_count+=1

print(data[i])

print("local minimum count:",local_minimum_count)

#source code along with output:

1 N 3 Ою No Ол в 7 8 data=(3,2,9,8,7,8,6] local_minimum_count=0 for i in range(1, len (data)-1): if(data[i-1]>data[i] and data[i+1]>data[i]): local_minimum_count+=1 print(data[i]) print("local minimum count:", local_minimum_count) Terminal File Edit View Search Terminal Help user@user-Latitude - 3490:-/Desktops python3.7 1.py 2 7 local minimum count: 2 user@user-Latitude - 3490:-/Desktop

#using file

data.txt

3
2
9
8
7
8
6

1 2 3 4 5 6 3 2 9 8 7 8 6 7

#source code:

f=open("data.txt","r")

data=f.read().split("\n")

data=[int(i) for i in data]

local_minimum_count=0

for i in range(1,len(data)-1):

if(data[i-1]>data[i] and data[i+1]>data[i]):

local_minimum_count+=1

print(data[i])

print("local minimum count:",local_minimum_count)

1 2 3 4. 5 6 7 8 9 10 f=open("data.txt","r") data=f.read().split("\n") data=[int(i) for i in data] local_minimum_count=0 for i in range(1, len(data)-1): if(data[i-1]>data[i] and data[i+1]>data[i]): local_minimum_count+=1 print (data[i]) print("local minimum count:", local_minimum_count) Terminal File Edit View Search Terminal Help user@user-Latitude - 3490:-/Desktop$ python3.7 1.py 2 7 local minimum count: 2 user@user-Latitude - 3490:-/Desktops I

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