# Python Challenge Answers Challenges for 10-26-18 - python answers  # cumulative sum x=[3,10,4,12,55] cs=[0]*5 for i in range(0,len(x)): cs[i]=sum(x[0:(i+1)]) print(cs) x=[3,10,4,12,55] cs=list() for i in range(0,len(x)): cs.append(sum(x[0:(i+1)])) print(cs) #how long to 5 5's x=[5,3,2,5,5,1,2,5,3,5,1,5,1] count=0 i=0 while count < 5: if x[i]==5: count+=1 i+=1 print(i) # answers for Lecture 11 extra practice import pandas as pd # 0 - calculate sum of female and male wages in wages.csv wages=pandas.read_csv("wages.csv",header=0,sep=",") femaleSum=0 maleSum=0 for i in range(0,len(wages),1): if wages.gender[i]=="female": femaleSum=femaleSum+wages.wage[i] else: maleSum=maleSum+wages.wage[i] femaleSum maleSum sum(wages.gender=="female") sum(wages.gender=="male") wages.tail() # find runs # load file findRuns=pd.read_csv("findRuns.txt",header=None,sep="\t") # create a variable out that is currently undefined out=pd.DataFrame(columns=['startIndex','runLength']) # I will use this variable cur to hold onto the previous number in the vector; # this is analagous to using findRuns[i-1] cur=findRuns.iloc[0,0] #cur=findRuns[i-1] # this is a counter that I use to keep track of how long a run of repeated values is; # if there are not repeated values than this count equals 1 count=1 # loop through each entry of our vector (except the 1st one, which we set to cur above) for i in range(1,50,1): # test if the ith value in the vector findRuns equals the previous (stored in cur) if findRuns.iloc[i,0]==cur: # test whether count is 1 (we aren't in the middle of a run) or >1 (in the middle of a run) if count==1: # if the ith value in the vector equals the previous (stored in cur) and count is 1, we # are at the beginning of a run and we want to store this value (we temporarily store it in 'start') start=(i-1) # we add one to count because the run continued based on the ith value of findRuns being equal to # the previous (stored in cur) count=count+1 # if the ith value in findRuns is not the same as the previous (stored in cur) we either are not in a run # or we are ending a run else: # if count is greater than 1 it means we were in a run and must be exiting one if count>1: # add a row to 'out' that will hold the starting positions in the first column and the length # of runs in the second column; this appends rows to out after finding and counting each run out.loc[len(out)]=[start,count] # reset count to 1 because we just exited a run count=1 # remember cur holds the previous element in findRuns, so we need to update this after each time # we go through the for loop cur=findRuns.iloc[i,0] cur out