A dummy dataset is :
data <- data.frame(
group = c(1,1,1,1,1,2),
dates = as.Date(c("2005-01-01", "2006-05-01", "2007-05-01","2004-08-01",
"2005-03-01","2010-02-01")),
value = c(10,20,NA,40,NA,5)
)
For each group, the missing values need to be filled with the non-missing value corresponding to the nearest date within same group. In case of a tie, pick any.
I am using dplyr. which.closest from birk but it needs a vector and a value. How to look up within a vector without writing loops. Even if there is an SQL solution, will do.
Any pointers to the solution?
May be something like: value = value[match(which.closest(dates,THISdate) & !is.na(value))]
Not sure how to specify Thisdate.
Edit: The expected value vector should look like:
value = c(10,20,20,40,10,5)
Using knn1 (nearest neighbor) from the class package (which comes with R -- don't need to install it) and dplyr define an na.knn1 function which replaces each NA value in x with the non-NA x value having the closest time.
library(class)
na.knn1 <- function(x, time) {
is_na <- is.na(x)
if (sum(is_na) == 0 || all(is_na)) return(x)
train <- matrix(time[!is_na])
test <- matrix(time[is_na])
cl <- x[!is_na]
x[is_na] <- as.numeric(as.character(knn1(train, test, cl)))
x
}
data %>% mutate(value = na.knn1(value, dates))
giving:
group dates value
1 1 2005-01-01 10
2 1 2006-05-01 20
3 1 2007-05-01 20
4 1 2004-08-01 40
5 1 2005-03-01 10
6 2 2010-02-01 5
Add an appropriate group_by if the intention was to do this by group.
You can try the use of sapply to find the values closest since the x argument in `which.closest only takes a single value.
first create a vect whereby the dates with no values are replaced with NA and use it within the which.closest function.
library(birk)
vect=replace(data$dates,which(is.na(data$value)),NA)
transform(data,value=value[sapply(dates,which.closest,vec=vect)])
group dates value
1 1 2005-01-01 10
2 1 2006-05-01 20
3 1 2007-05-01 20
4 1 2004-08-01 40
5 1 2005-03-01 10
6 2 2010-02-01 5
if which.closest was to take a vector then there would be no need of sapply. But this is not the case.
Using the dplyr package:
library(birk)
library(dplyr)
data%>%mutate(vect=`is.na<-`(dates,is.na(value)),
value=value[sapply(dates,which.closest,vect)])%>%
select(-vect)
Related
Pandas help!
I have a specific column like this,
Mpg
0 18
1 17
2 19
3 21
4 16
5 15
Mpg is mile per gallon,
Now I need to replace that 'MPG' column to 'litre per 100 km' and change those values to litre per 100 km' at the same time. Any help? Thanks beforehand.
-Tom
I changed the name of the column but doing both simultaneously,i could not.
Use pop to return and delete the column at the same time and rdiv to perform the conversion (1 mpg = 1/235.15 liter/100km):
df['litre per 100 km'] = df.pop('Mpg').rdiv(235.15)
If you want to insert the column in the same position:
df.insert(df.columns.get_loc('Mpg'), 'litre per 100 km',
df.pop('Mpg').rdiv(235.15))
Output:
litre per 100 km
0 13.063889
1 13.832353
2 12.376316
3 11.197619
4 14.696875
5 15.676667
An alternative to pop would be to store the result in another dataframe. This way you can perform the two steps at the same time. In my code below, I first reproduce your dataframe, then store the constant for conversion and perform it on all entries using the apply method.
df = pd.DataFrame({'Mpg':[18,17,19,21,16,15]})
cc = 235.214583 # constant for conversion from mpg to L/100km
df2 = pd.DataFrame()
df2['litre per 100 km'] = df['Mpg'].apply(lambda x: cc/x)
print(df2)
The output of this code is:
litre per 100 km
0 13.067477
1 13.836152
2 12.379715
3 11.200694
4 14.700911
5 15.680972
as expected.
My code uses a column called booking status that is 1 for yes and 0 for no (there are multiple other columns that information will be pulled from dependant on the booking status) - there are lots more no than yes so I would like to take a sample with all the yes and the same amount of no.
When I use
samp = rslt_df.sample(n=298, random_state=1, weights='bookingstatus')
I get the error:
ValueError: Fewer non-zero entries in p than size
Is there a way to do this sample this way?
If our entire dataset looks like this:
print(df)
c1 c2
0 1 1
1 0 2
2 0 3
3 0 4
4 0 5
5 0 6
6 0 7
7 1 8
8 0 9
9 0 10
We may decide to sample from it using the DataFrame.sample function. By default, this function will sample without replacement. Meaning, you'll receive an error by specifying a number of observations larger than the number of observations in your initial dataset:
df.sample(20)
ValueError: Cannot take a larger sample than population when 'replace=False'
In your situation, the ValueError comes from the weights parameter:
df.sample(3,weights='c1')
ValueError: Fewer non-zero entries in p than size
To paraphrase the DataFrame.sample docs, using the c1 column as our weights parameter implies that rows with a larger value in the c1 column are more likely to be sampled. Specifically, the sample function will not pick values from this column that are zero. We can fix this error using either one of the following methods.
Method 1: Set the replace parameter to be true:
m1 = df.sample(3,weights='c1', replace=True)
print(m1)
c1 c2
0 1 1
7 1 8
0 1 1
Method 2: Make sure the n parameter is equal to or less than the number of 1s in the c1 column:
m2 = df.sample(2,weights='c1')
print(m2)
c1 c2
7 1 8
0 1 1
If you decide to use this method, you won't really be sampling. You're really just filtering out any rows where the value of c1 is 0.
I was able to this in the end, here is how I did it:
bookingstatus_count = df.bookingstatus.value_counts()
print('Class 0:', bookingstatus_count[0])
print('Class 1:', bookingstatus_count[1])
print('Proportion:', round(bookingstatus_count[0] / bookingstatus_count[1], 2), ': 1')
# Class count
count_class_0, count_class_1 = df.bookingstatus.value_counts()
# Divide by class
df_class_0 = df[df['bookingstatus'] == 0]
df_class_0_under = df_class_0.sample(count_class_1)
df_test_under = pd.concat([f_class_0_under, df_class_1], axis=0)
df_class_1 = df[df['bookingstatus'] == 1]
based on this https://www.kaggle.com/rafjaa/resampling-strategies-for-imbalanced-datasets
Thanks everyone
I have a data frame with the following two variables:
amount: num 1213.5 34.5 ...
txn_date: POSIXct, format "2017-05-01 12:13:30" ...
I want to transform it in a time series using ts().
I started using this code:
Z <- zoo(data$amount, order.by=as.Date(as.character(data$txn_date), format="%Y/%m/%d %H:%M:%S"))
But the problem is that in Z I loose the dates. In fact, all the dates are reported as NA.
How can I solve it?
For my analysis is important to have date in the format:%Y/%m/%d %H:%M:%S
for example 2017-05-01 12:13:30. I don't want to remove the time component in the variable txn_date.
Yhan you for your help,
Andrea
I think your prolem comes from the way you're manipulating your data frame, could post more details about it please ?
I think i have a fix for you.
Data frame I used :
> df1
$data
value
1 1.9150
2 3.1025
3 6.7400
4 8.5025
5 11.0025
6 9.8025
7 9.0775
8 7.0900
9 6.8525
10 7.4900
$date
%Y-%m-%d
1 1974-01-01
2 1974-01-02
3 1974-01-03
4 1974-01-04
5 1974-01-05
6 1974-01-06
7 1974-01-07
8 1974-01-08
9 1974-01-09
10 1974-01-10
> class(df1$data$value)
[1] "numeric"
> class(df1$date$`%Y-%m-%d`)
[1] "POSIXct" "POSIXt"
Then I can create a time serie by calling zoo like that :
> Z<-zoo(df1$data,order.by=(as.POSIXct(df1$date$`%Y-%m-%d`)))
> Z
value
1974-01-01 1.9150
1974-01-02 3.1025
1974-01-03 6.7400
1974-01-04 8.5025
1974-01-05 11.0025
1974-01-06 9.8025
1974-01-07 9.0775
1974-01-08 7.0900
1974-01-09 6.8525
1974-01-10 7.4900
The important thing here is that I use df1$date$%Y-%m-%d instead of just
df1$date
In fact if I try the way you did it I get NA values too :
> Z<-zoo(df1$data,order.by=as.POSIXct(as.Date(as.character(df1$date),format("%Y-%m-%d"))))
> Z
value
<NA> 1.915
To get the name of data$txn_date you can use the following command : names(data$txn_date) and try my solution with your data frame and name.
> names(df1$date)
[1] "%Y-%m-%d"
The rolling function in Pandas can only calculate rolling statistics according to row counts or date/time columns. But I want to have a discrete time column for calculating rolling sum, something like this:
key time value
A 1 10
A 2 20
A 4 30
A 7 10
B 1 15
B 2 30
B 3 15
I want to first group by key, then calculate the rolling sum on value for the nearest 3 time:
key time value output
A 1 10 10
A 2 20 30(10+20)
A 4 30 60(10+20+30)
A 7 10 40(30+10)
B 1 15 15
B 2 30 45
B 3 15 60
I tried this:
grouped = input.groupby("key", as_index=False)
for name, group in grouped:
group = group.sort_values("time")
time = list(group["time"])
value = list(group["value"])
#calcRollingStat is a custom function that outputs a list of corresponding results
out = calcRollingStat(time, value, mode="avg")
group["output"] = out #out is a list
But then I don't know how to convert grouped back to DataFrame. Pandas tells me that there is no reset_index attribute in grouped.
Is my code the best method to do this? How would you tackle this problem?
Thank you!
I believe you can use GroupBy.apply with custom function:
def f(group):
group = group.sort_values("time")
time = list(group["time"])
value = list(group["value"])
#calcRollingStat is a custom function that outputs a list of corresponding results
group["output"] = calcRollingStat(time, value, mode="avg")
return group
df = input.groupby("key", as_index=False).apply(f)
From my "Id" Column I want to remove the one and zero's from the left.
That is
1000003 becomes 3
1000005 becomes 5
1000011 becomes 11 and so on
Ignore -1, 10 and 1000000, they will be handled as special cases. but from the remaining rows I want to remove the "1" followed by zeros.
Well you can use modulus to get the end of the numbers (they will be the remainder). So just exclude the rows with ids of [-1,10,1000000] and then compute the modulus of 1000000:
print df
Id
0 -1
1 10
2 1000000
3 1000003
4 1000005
5 1000007
6 1000009
7 1000011
keep = df.Id.isin([-1,10,1000000])
df.Id[~keep] = df.Id[~keep] % 1000000
print df
Id
0 -1
1 10
2 1000000
3 3
4 5
5 7
6 9
7 11
Edit: Here is a fully vectorized string slice version as an alternative (Like Alex' method but takes advantage of pandas' vectorized string methods):
keep = df.Id.isin([-1,10,1000000])
df.Id[~keep] = df.Id[~keep].astype(str).str[1:].astype(int)
print df
Id
0 -1
1 10
2 1000000
3 3
4 5
5 7
6 9
7 11
Here is another way you could try to do it:
def f(x):
"""convert the value to a string, then select only the characters
after the first one in the string, which is 1. For example,
100005 would be 00005 and I believe it's returning 00005.0 from
dataframe, which is why the float() is there. Then just convert
it to an int, and you'll have 5, etc.
"""
return int(float(str(x)[1:]))
# apply the function "f" to the dataframe and pass in the column 'Id'
df.apply(lambda row: f(row['Id']), axis=1)
I get that this question is satisfactory answered. But for future visitors, what I like about alex' answer is that it does not depend on there to be exactly four zeros. The accepted answer will fail if you sometimes have 10005, sometimes 1000005 and whatever.
However, to add something more to the way we think about it. If you know it's always going to be 10000, you can do
# backup all values
foo = df.id
#now, some will be negative or zero
df.id = df.id - 10000
#back in those that are negative or zero (here, first three rows)
df.if[df.if <= 0] = foo[df.id <= 0]
It gives you the same as Karl's answer, but I typically prefer these kind of methods for their readability.