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By : Loki
Date : November 21 2020, 04:01 AM
will help you Hi I have a bunch of data and I need to know how much of each value show up and how often the frequency is only corresponding to the unique value. I used a countif to count the unique value but I wanna know the frequency of the amount of times it shows up. ,
Change the formula in B1 to, code :
``````=IF(COUNTIF(A\$1:A1, A1)=1, COUNTIF(A:A, A1), TEXT(,))
``````
``````=COUNTIF(B:B, E1)
`````` ## Python Excel number of times a negative and positive number appears (count/frequency)

By : user3363273
Date : March 29 2020, 07:55 AM
it fixes the issue Use two counter variables to track of the total count and number of positives. Set them to 0 in the beginning, and then inside of your loop, increment them by using += 1 whenever you want to add 1.
Then test whether the percentage is greater than 0 by stripping out the percentage sign and then converting the string into a number using if float(row.strip('%')) > 0. You can change this to >= if you want to include 0 in the "positive" category.
code :
``````totalCount = 0
numberOfPositives = 0

with open('FL%.csv', 'r') as file2:
reader.next() # this skips the first row of the file
# this iteration will start from the second row of file2.csv
conditionMet = False
if conditionMet == True:
if float(row.strip('%')) > 0: # change > to >= if you want to count 0 as positive
print "FA, 1",row,',', ','.join(row[1:5]) # print 1 if positive
numberOfPositives += 1 # add 1 to numberOfPositives only if positive
else:
print "FA, 0",row,',', ','.join(row[1:5]) # print 0 if not positive
totalCount += 1 # add 1 to totalCount regardless of sign
conditionMet = False # or break if you know you only need at most one line
if row[1:5] == val1:
conditionMet = True
``````
``````print 'Total Count =', totalCount
print 'Percentage of Positive numbers =', numberOfPositives * 100./totalCount, '%'
`````` ## Python count the frequency of each number in a list without any count() functions or without importing anything

By : user3408415
Date : March 29 2020, 07:55 AM
Any of those help Lets say I have List myList = [1, 1, 2, 2, 2, 3] I want to count the frequency of each number in the list and make it into some sort of graph with an output: , A dict-based solution:
code :
``````myList = [1, 1, 2, 2, 2, 3]
output = {}

for item in myList:
if item not in output:
output[item] = 0
output[item] += 1
``````
``````for number, count in output.iteritems():
print "{0}: {1}".format(number, "X " * count)
`````` ## Count number of cells that match criteria with nested IF and COUNTIF statements

By : user3098577
Date : March 29 2020, 07:55 AM
it should still fix some issue I'm trying to use an excel formula to count the number of cells in a given range (let's say T132:T134) that match a specific line of text, however cells should only be counted if in a different range (say F132:F134) the corresponding cell in the same row is different from a specific line of text. Here is what I am trying to do: , Didnt you meant to use COUNTIFS?:
code :
``````=COUNTIFS(T132:T134,"<>Not applied",F132:F134,"WN(1)")
``````
``````=COUNTIFS(T132:T134,"<>Not applied",F132:F134,"*WN(1)*")
`````` ## Using only an Excel formula, how do I count the number of rows that meet multiple criteria (including a countif)?

By : user3642527
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I have a spreadsheet with 10 columns. It contains the name of an item, 8 different ratings, and a dollar amount. If there is not a rating for a specific column, the value is "NR" (for not rated). See attached example. , In J2:
code :
``````=SUM(--(MMULT(--(B7:I16<>"NR"),TRANSPOSE(COLUMN(B7:I16)^0))=1))
``````
``````=SUM((MMULT(--(B7:I16<>"NR"),TRANSPOSE(COLUMN(B7:I16)^0))=1)*J7:J16)
`````` ## Count past relative number of rows that meet criteria similar to excel's COUNTIF

By : user3670979
Date : March 29 2020, 07:55 AM
will help you I have a stock data csv with the following info , IIUC:
code :
``````df["above"] = pd.DataFrame([df["High"].shift(-1)>df["Open"],
df["High"].shift(-2)>df["Open"]]).T.sum(axis=1)

# or [df["High"].shift(n)>df["Open"] for n in range(-1,-5,-1)] if you want to generalize the number n

print (df)

Open    High     Low   Close  above
0  154.55  155.54  152.90  153.41      2
1  156.82  158.75  155.42  156.76      1
2  150.21  157.44  150.15  156.33      0
3  147.78  149.38  146.88  149.11      0
4  144.25  147.28  143.90  146.27      0
5  142.90  144.05  140.79  143.73      0
`````` 