How to Count the Number of Unique Values by Group

Count unique values using pandas groupby

I think you can use SeriesGroupBy.nunique:

print (df.groupby('param')['group'].nunique())
param
a    2
b    1
Name: group, dtype: int64

Another solution with unique, then create new df by DataFrame.from_records, reshape to Series by stack and last value_counts:

a = df[df.param.notnull()].groupby('group')['param'].unique()
print (pd.DataFrame.from_records(a.values.tolist()).stack().value_counts())
a    2
b    1
dtype: int64

Add count of unique / distinct values by group to the original data

Using ave (since you ask for it specifically):

within(df, { count <- ave(type, color, FUN=function(x) length(unique(x)))})

Make sure that type is character vector and not factor.

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Since you also say your data is huge and that speed/performance may therefore be a factor, I'd suggest a data.table solution as well.

require(data.table)
setDT(df)[, count := uniqueN(type), by = color] # v1.9.6+
# if you don't want df to be modified by reference
ans = as.data.table(df)[, count := uniqueN(type), by = color]

uniqueN was implemented in v1.9.6 and is a faster equivalent of length(unique(.)). In addition it also works with data.frames/data.tables.

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Other solutions:

Using plyr:

require(plyr)
ddply(df, .(color), mutate, count = length(unique(type)))

Using aggregate:

agg <- aggregate(data=df, type ~ color, function(x) length(unique(x)))
merge(df, agg, by="color", all=TRUE)

Count unique values per groups with Pandas

You need nunique:

df = df.groupby('domain')['ID'].nunique()

print (df)
domain
'facebook.com'    1
'google.com'      1
'twitter.com'     2
'vk.com'          3
Name: ID, dtype: int64

If you need to strip ' characters:

df = df.ID.groupby([df.domain.str.strip("'")]).nunique()
print (df)
domain
facebook.com    1
google.com      1
twitter.com     2
vk.com          3
Name: ID, dtype: int64

Or as Jon Clements commented:

df.groupby(df.domain.str.strip("'"))['ID'].nunique()

You can retain the column name like this:

df = df.groupby(by='domain', as_index=False).agg({'ID': pd.Series.nunique})
print(df)
    domain  ID
0       fb   1
1      ggl   1
2  twitter   2
3       vk   3

The difference is that nunique() returns a Series and agg() returns a DataFrame.

How to count the number of unique values by group?

I think you've got it all wrong here. There is no need neither in plyr or <- when using data.table.

Recent versions of data.table, v >= 1.9.6, have a new function uniqueN() just for that.

library(data.table) ## >= v1.9.6
setDT(d)[, .(count = uniqueN(color)), by = ID]
#    ID count
# 1:  A     3
# 2:  B     2

If you want to create a new column with the counts, use the := operator

setDT(d)[, count := uniqueN(color), by = ID]

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Or with dplyr use the n_distinct function

library(dplyr)
d %>%
  group_by(ID) %>%
  summarise(count = n_distinct(color))
# Source: local data table [2 x 2]
# 
#   ID count
# 1  A     3
# 2  B     2

Or (if you want a new column) use mutate instead of summary

d %>%
  group_by(ID) %>%
  mutate(count = n_distinct(color))

Count distinct values depending on group

You would use count(distinct):

select "group", count(distinct id)
from t
group by "group";

Note that group is a very poor name for a column because it is a SQL keyword. Hopefully the real column name is something more reasonable.

Selecting COUNT(*) with DISTINCT

Count all the DISTINCT program names by program type and push number

SELECT COUNT(DISTINCT program_name) AS Count,
  program_type AS [Type] 
FROM cm_production 
WHERE push_number=@push_number 
GROUP BY program_type

DISTINCT COUNT(*) will return a row for each unique count. What you want is COUNT(DISTINCT <expression>): evaluates expression for each row in a group and returns the number of unique, non-null values.

Count the number of unique values per group

Looks like you want transform + nunique;

df['a_b_3'] = df.groupby('_a')['_b'].transform('nunique')        
df
   _a  _b  a_b_3
0   1   3      3
1   1   4      3
2   1   5      3
3   2   3      1
4   2   3      1
5   3   3      2
6   3   9      2

This is effectively groupby + nunique + map:

v = df.groupby('_a')['_b'].nunique()
df['a_b_3'] = df['_a'].map(v)

df
   _a  _b  a_b_3
0   1   3      3
1   1   4      3
2   1   5      3
3   2   3      1
4   2   3      1
5   3   3      2
6   3   9      2

How to count number of unique groups missing information in a groupby?

To count unique IDs, check where it's null then max within [ID, method], to indicate any missing value within that [ID, method]. Then sum over the method to get the Number of unique IDS missing something.

(df[['var_1', 'var_2']].isnull()
    .groupby([df['ID'], df['method']]).max()
    .sum(level='method')

---

        var_1  var_2
method              
AB          1      0
CD          1      0
BC          1      0
DE          0      1

Python group by and count distinct values in a column and create delimited list

You can use str.len in your code:

df3 = (df.groupby('company')['product']
         .apply(lambda x: list(x.unique()))
         .reset_index()
         .assign(count=lambda d: d['product'].str.len())  ## added line
      )

output:

     company            product  count
0     Amazon           [E-comm]      1
1   Facebook     [Social Media]      1
2     Google  [Search, Android]      2
3  Microsoft        [OS, X-box]      2

R - Count unique/distinct values in two columns together per group

You can subset the data from cur_data() and unlist the data to get a vector. Use n_distinct to count number of unique values.

library(dplyr)

df %>%
  group_by(ID) %>%
  mutate(Count = n_distinct(unlist(select(cur_data(), 
                   Party, Party2013)), na.rm = TRUE)) %>%
  ungroup


#     ID  Wave Party Party2013 Count
#  <int> <int> <chr> <chr>     <int>
#1     1     1 A     A             2
#2     1     2 A     NA            2
#3     1     3 B     NA            2
#4     1     4 B     NA            2
#5     2     1 A     C             3
#6     2     2 B     NA            3
#7     2     3 B     NA            3
#8     2     4 B     NA            3

data

It is easier to help if you provide data in a reproducible format

df <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), Wave = c(1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L), Party = c("A", "A", "B", "B", "A", 
"B", "B", "B"), Party2013 = c("A", NA, NA, NA, "C", NA, NA, NA
)), class = "data.frame", row.names = c(NA, -8L))

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