I have a dataframe with duplicated rows except one column value, I want to drop the row with a value of "None" if the id is the same (not all the rows are duplicated)
a b
1 1 None
2 1 7
3 2 2
4 3 4
I need to drop the first row with the duplicated (1) and the value of b is None.
You can use duplicated and also search for None. That will return the row you want to drop, so use ~ to get the inverse dataframe (so everything but the row you want to drop) to return the expected result. EDIT: Passing keep=False will return all duplicates, so order doesn't matter.
df[~((df['b'].isnull()) & (df.duplicated('a', keep=False)))] #if None is Null value
OR
df[~((df['b'] == 'None') & (df.duplicated('a', keep=False)))] if 'None' is string
I'm rather new to pandas and recently run into a problem. I have a pandas DataFrame that I need to process. I need to extract parts of the DataFrame where specific conditions are met. However, i want these parts to be coherent blocks, not one big set.
Example:
Consider the following pandas DataFrame
col1 col2
0 3 11
1 7 15
2 9 1
3 11 2
4 13 2
5 16 16
6 19 17
7 23 13
8 27 4
9 32 3
I want to extract the subframes where the values of col2 >= 10, resulting maybe in a list of DataFrames in the form of (in this case):
col1 col2
0 3 11
1 7 15
col1 col2
5 16 16
6 19 17
7 23 13
Ultimately, I need to do further analysis on the values in col1 within the resulting parts. However, the start and end of each of these blocks is important to me, so simply creating a subset using pandas.DataFrame.loc isn't going to work for me, i think.
What I have tried:
Right now I have a workaround that gets the subset using pandas.DataFrame.loc and then extracts the start and end index of each coherent block afterwards, by iterating through the subset and check, whether there is a jump in the indices. However, it feels rather clumsy and I feel that I'm missing a basic pandas function here, that would make my code more efficient and clean.
This is code representing my current workaround as adapted to the above example
# here the blocks will be collected for further computations
blocks = []
# get all the items where col2 >10 using 'loc[]'
subset = df.loc[df['col2']>10]
block_start = 0
block_end = None
#loop through all items in subset
for i in range(1, len(subset)):
# if the difference between the current index and the last is greater than 1 ...
if subset.index[i]-subset.index[i-1] > 1:
# ... this is the current blocks end
next_block_start = i
# extract the according block and add it to the list of all blocks
block = subset[block_start:next_block_start]
blocks.append(block)
#the next_block_start index is now the new block's starting index
block_start = next_block_start
#close and add last block
blocks.append(subset[block_start:])
Edit: I was by mistake previously referring to 'pandas.DataFrame.where' instead of 'pandas.DataFrame.loc'. I seem to be a bit confused by my recent research.
You can split you problem into parts. At first you check the condition:
df['mask'] = (df['col2']>10)
We use this to see where a new subset starts:
df['new'] = df['mask'].gt(df['mask'].shift(fill_value=False))
Now you can combine these informations into a group number. The cumsum will generate a step function which we set to zero (via the mask column) if this is not a group we are interested in.
df['grp'] = (df.new + 0).cumsum() * df['mask']
EDIT
You don't have to do the group calculation in your df:
s = (df['col2']>10)
s = (s.gt(s.shift(fill_value=False)) + 0).cumsum() * s
After that you can split this into a dict of separate DataFrames
grp = {}
for i in np.unique(s)[1:]:
grp[i] = df.loc[s == i, ['col1', 'col2']]
I have a log file per day which is placed in my LAN on a http server that grows to about 3MB per day.
Every 15 seconds new values are written to that file. It has a timestamp column. There are many other columns which are not needed for me, so I only need about 5 of the columns.
Pandas should "monitor" that file by reading only records which are new. Let's say last execution was 2018-02-05 00:00:04.467 then this should be the filter for next runtime (>2018-02-05 00:00:04.467) and in the end of this runtime the last timestamp read should be filter for next and so on...
I'm new to pandas and haven't found any similar thread for this.
I guess the CSV would be written line by line, so instead of reading the whole file and filtering, you could accumulate the number of rows in the file in a variable rows and for the next run, use read_csv passing in the optional argument skiprows with value range(1, rows + 1) to skip the first rows in the file, and then incrementing rows += len(df)
If data.csv is
a,b,c
1,2,3
4,5,6
7,8,9
3,2,1
6,5,4
and rows = 2 (i.e., the last time the file was read it had 2 rows) then
df = pd.read_csv("data.csv", usecols=["a", "c"], skiprows=range(1, rows + 1))
would be the dataframe
a c
0 7 9
1 3 1
2 6 4
and you would increment rows
rows += len(df) # rows now equals 5, so 5 rows would be skipped in the next run
I have got a data set that contains 3 columns and has 15565 observations. one of the columns has got several words in the same row.
What I am looking to do is to extract a particular word from each row and append it to a new column (i will have 4 cols in total)
The problem is that the word that i am looking for are not the same and they are not always on the same position.
Here is an extract of my DS:
x y z
-----------------------------------------------------------------------
1 T 3C00652722 (T558799A)
2 T NA >> MSP: T0578836A & 3C03024632
3 T T0579010A, 3C03051500, EAET03051496
4 U T0023231A > MSP: T0577506A & 3C02808556
8 U (T561041A C72/59460)>POPMigr.T576447A,C72/221816*3C00721502
I am looking to extract all the words that start with 3Cand are 10 characters long and then append the to a new col so it looks like this:
x y z Ref
----------------------------------------------------------------
1 T 3C00652722 (T558799A) 3C00652722
2 T NA >> MSP: T0578836A & 3C03024632 3C03024632
3 T T0579010A, 3C03051500, EAET03051496 3C03051500
4 U T0023231A > MSP: T0577506A & 3C02808556 3C02808556
8 U >POPMigr.T576447A,C72/221816*3C00721502 3C00721502
I have tried using the Contains, Like and substring methods but it does not give me the results i am looking for as it basically finds the rows that have the 3C number but does not extract it, it just copies the whole cell and pastes is on the Ref column.
SQL Server doesn't have good string functions, but this should suffice if you only want to extract one value per row:
select t.*,
left(stuff(col,
1,
patindex('%3C[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]%', col),
''
), 10)
from t ;
I'm trying to create 20 unique cards with numbers, but I struggle a bit.. So basically I need to create 20 unique matrices 3x3 having numbers 1-10 in first column, numbers 11-20 in the second column and 21-30 in the third column.. Any ideas? I'd prefer to have it done in r, especially as I don't know Visual Basic. In excel I know how to generate the cards, but not sure how to ensure they are unique..
It seems to be quite precise and straightforward to me. Anyway, i needed to create 20 matrices that would look like :
[,1] [,2] [,3]
[1,] 5 17 23
[2,] 8 18 22
[3,] 3 16 24
Each of the matrices should be unique and each of the columns should consist of three unique numbers ( the 1st column - numbers 1-10, the 2nd column 11-20, the 3rd column - 21-30).
Generating random numbers is easy, though how to make sure that generated cards are unique?Please have a look at the post that i voted for as an answer - as it gives you thorough explanation how to achieve it.
(N.B. : I misread "rows" instead of "columns", so the following code and explanation will deal with matrices with random numbers 1-10 on 1st row, 11-20 on 2nd row etc., instead of columns, but it's exactly the same just transposed)
This code should guarantee uniqueness and good randomness :
library(gtools)
# helper function
getKthPermWithRep <- function(k,n,r){
k <- k - 1
if(n^r< k){
stop('k is greater than possibile permutations')
}
v <- rep.int(0,r)
index <- length(v)
while ( k != 0 )
{
remainder<- k %% n
k <- k %/% n
v[index] <- remainder
index <- index - 1
}
return(v+1)
}
# get all possible permutations of 10 elements taken 3 at a time
# (singlerowperms = 720)
allperms <- permutations(10,3)
singlerowperms <- nrow(allperms)
# get 20 random and unique bingo cards
cards <- lapply(sample.int(singlerowperms^3,20),FUN=function(k){
perm2use <- getKthPermWithRep(k,singlerowperms,3)
m <- allperms[perm2use,]
m[2,] <- m[2,] + 10
m[3,] <- m[3,] + 20
return(m)
# if you want transpose the result just do:
# return(t(m))
})
Explanation
(disclaimer tl;dr)
To guarantee both randomness and uniqueness, one safe approach is generating all the possibile bingo cards and then choose randomly among them without replacements.
To generate all the possible cards, we should :
generate all the possibilities for each row of 3 elements
get the cartesian product of them
Step (1) can be easily obtained using function permutations of package gtools (see the object allPerms in the code). Note that we just need the permutations for the first row (i.e. 3 elements taken from 1-10) since the permutations of the other rows can be easily obtained from the first by adding 10 and 20 respectively.
Step (2) is also easy to get in R, but let's first consider how many possibilities will be generated. Step (1) returned 720 cases for each row, so, in the end we will have 720*720*720 = 720^3 = 373248000 possible bingo cards!
Generate all of them is not practical since the occupied memory would be huge, thus we need to find a way to get 20 random elements in this big range of possibilities without actually keeping them in memory.
The solution comes from the function getKthPermWithRep, which, given an index k, it returns the k-th permutation with repetition of r elements taken from 1:n (note that in this case permutation with repetition corresponds to the cartesian product).
e.g.
# all permutations with repetition of 2 elements in 1:3 are
permutations(n = 3, r = 2,repeats.allowed = TRUE)
# [,1] [,2]
# [1,] 1 1
# [2,] 1 2
# [3,] 1 3
# [4,] 2 1
# [5,] 2 2
# [6,] 2 3
# [7,] 3 1
# [8,] 3 2
# [9,] 3 3
# using the getKthPermWithRep you can get directly the k-th permutation you want :
getKthPermWithRep(k=4,n=3,r=2)
# [1] 2 1
getKthPermWithRep(k=8,n=3,r=2)
# [1] 3 2
Hence now we just choose 20 random indexes in the range 1:720^3 (using sample.int function), then for each of them we get the corresponding permutation of 3 numbers taken from 1:720 using function getKthPermWithRep.
Finally these triplets of numbers, can be converted to actual card rows by using them as indexes to subset allPerms and get our final matrix (after, of course, adding +10 and +20 to the 2nd and 3rd row).
Bonus
Explanation of getKthPermWithRep
If you look at the example above (permutations with repetition of 2 elements in 1:3), and subtract 1 to all number of the results you get this :
> permutations(n = 3, r = 2,repeats.allowed = T) - 1
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 0 2
[4,] 1 0
[5,] 1 1
[6,] 1 2
[7,] 2 0
[8,] 2 1
[9,] 2 2
If you consider each number of each row as a number digit, you can notice that those rows (00, 01, 02...) are all the numbers from 0 to 8, represented in base 3 (yes, 3 as n). So, when you ask the k-th permutation with repetition of r elements in 1:n, you are also asking to translate k-1 into base n and return the digits increased by 1.
Therefore, given the algorithm to change any number from base 10 to base n :
changeBase <- function(num,base){
v <- NULL
while ( num != 0 )
{
remainder = num %% base # assume K > 1
num = num %/% base # integer division
v <- c(remainder,v)
}
if(is.null(v)){
return(0)
}
return(v)
}
you can easily obtain getKthPermWithRep function.
One 3x3 matrix with the desired value range can be generated with the following code:
mat <- matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30, 3)), nrow=3)
Furthermore, you can use a for loop to generate a list of 20 unique matrices as follows:
for (i in 1:20) {
mat[[i]] <- list(matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30,3)), nrow=3))
print(mat[[i]])
}
Well OK I may fall on my face here but I propose a checksum (using Excel).
This is a unique signature for each bingo card which will remain invariate if the order of numbers within any column is changed without changing the actual numbers. The formula is
=SUM(10^MOD(A2:A4,10)+2*10^MOD(B2:B4,10)+4*10^MOD(C2:C4,10))
where the bingo numbers for the first card are in A2:C4.
The idea is to generate a 10-digit number for each column, then multiply each by a constant and add them to get the signature.
So here I have generated two random bingo cards using a standard formula from here plus two which are deliberately made to be just permutations of each other.
Then I check if any of the signatures are duplicates using the formula
=MAX(COUNTIF(D5:D20,D5:D20))
which shouldn't given an answer more than 1.
In the unlikely event that there were duplicates, then you would just press F9 and generate some new cards.
All formulae are array formulae and must be entered with CtrlShiftEnter
Here is an inelegant way to do this. Generate all possible combinations and then sample without replacement. These are permutations, combinations: order does matter in bingo
library(dplyr)
library(tidyr)
library(magrittr)
generate_samples = function(n) {
first = data_frame(first = (n-9):n)
first %>%
merge(first %>% rename(second = first)) %>%
merge(first %>% rename(third = first)) %>%
sample_n(20)
}
suffix = function(df, suffix)
df %>%
setNames(names(.) %>%
paste0(suffix))
generate_samples(10) %>% suffix(10) %>%
bind_cols(generate_samples(20) %>% suffix(20)) %>%
bind_cols(generate_samples(30) %>% suffix(30)) %>%
rowwise %>%
do(matrix = t(.) %>% matrix(3)) %>%
use_series(matrix)