Mixing
How to assign count of unique values to the records in a data frame in python
8
I have a data frame like this:
IP_address
IP1
IP1
IP1
IP4
IP4
IP4
IP4
IP4
IP7
IP7
IP7
I would like to take count of unique values in this column and add the count as a variable by itself. At the end, it should look like this:
IP_address IP_address_Count
IP1 3
IP1 3
IP1 3
IP4 5
IP4 5
IP4 5
IP4 5
IP4 5
IP7 3
IP7 3
IP7 3
I am able to take the unique values of the column using the below code:
unique_ip_address_count = (df_c_train.drop_duplicates().IP_address.value_counts()).to_dict()
However, I am not sure how to match these in a loop in python so that i can get the desired results in python. Any sort of help is much appreciated.
I am not able to find a equivalent answer in stackoverflow. If there is anything please direct me there. Thank you.
python pandas
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
add a comment |
8
I have a data frame like this:
IP_address
IP1
IP1
IP1
IP4
IP4
IP4
IP4
IP4
IP7
IP7
IP7
I would like to take count of unique values in this column and add the count as a variable by itself. At the end, it should look like this:
IP_address IP_address_Count
IP1 3
IP1 3
IP1 3
IP4 5
IP4 5
IP4 5
IP4 5
IP4 5
IP7 3
IP7 3
IP7 3
I am able to take the unique values of the column using the below code:
unique_ip_address_count = (df_c_train.drop_duplicates().IP_address.value_counts()).to_dict()
However, I am not sure how to match these in a loop in python so that i can get the desired results in python. Any sort of help is much appreciated.
I am not able to find a equivalent answer in stackoverflow. If there is anything please direct me there. Thank you.
python pandas
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
add a comment |
8
8
8
5
I have a data frame like this:
IP_address
IP1
IP1
IP1
IP4
IP4
IP4
IP4
IP4
IP7
IP7
IP7
I would like to take count of unique values in this column and add the count as a variable by itself. At the end, it should look like this:
IP_address IP_address_Count
IP1 3
IP1 3
IP1 3
IP4 5
IP4 5
IP4 5
IP4 5
IP4 5
IP7 3
IP7 3
IP7 3
I am able to take the unique values of the column using the below code:
unique_ip_address_count = (df_c_train.drop_duplicates().IP_address.value_counts()).to_dict()
However, I am not sure how to match these in a loop in python so that i can get the desired results in python. Any sort of help is much appreciated.
I am not able to find a equivalent answer in stackoverflow. If there is anything please direct me there. Thank you.
python pandas
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
I have a data frame like this:
IP_address
IP1
IP1
IP1
IP4
IP4
IP4
IP4
IP4
IP7
IP7
IP7
I would like to take count of unique values in this column and add the count as a variable by itself. At the end, it should look like this:
IP_address IP_address_Count
IP1 3
IP1 3
IP1 3
IP4 5
IP4 5
IP4 5
IP4 5
IP4 5
IP7 3
IP7 3
IP7 3
I am able to take the unique values of the column using the below code:
unique_ip_address_count = (df_c_train.drop_duplicates().IP_address.value_counts()).to_dict()
However, I am not sure how to match these in a loop in python so that i can get the desired results in python. Any sort of help is much appreciated.
I am not able to find a equivalent answer in stackoverflow. If there is anything please direct me there. Thank you.
python pandas
python pandas
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
asked Sep 20 '17 at 20:25
Doubt DhanabaluDoubt Dhanabalu
1721313
1721313
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
10
You can use value_counts() with map
df['count'] = df['IP_address'].map(df['IP_address'].value_counts())
IP_address count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
|
show 3 more comments
9
Using pd.factorize
This should be a very fast solution that scales well for large data
f, u = pd.factorize(df.IP_address.values)
df.assign(IP_address_Count=np.bincount(f)[f])
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
add a comment |
8
NumPy way -
tags, C = np.unique(df.IP_address, return_counts=1, return_inverse=1)[1:]
df['IP_address_Count'] = C[tags]
Sample output -
In [275]: df
Out[275]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
add a comment |
7
In [75]: df['IP_address_Count'] = df.groupby('IP_address')['IP_address'].transform('size')
In [76]: df
Out[76]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
add a comment |
1
ip_set = df.IP_address.unique()
dict_temp = {}
for ip in ip_set:
dict_temp[ip] = df[df.IP_address == ip].IP_address.value_counts()[0]
df['counts'] = [dict_temp[ip] for ip in df.IP_address]
This seems to give me the sort of output that you desire
EDIT: Vaishali's use of map is perfect
answered Sep 20 '17 at 20:41
NRKNRK
363
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
10
You can use value_counts() with map
df['count'] = df['IP_address'].map(df['IP_address'].value_counts())
IP_address count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
|
show 3 more comments
10
You can use value_counts() with map
df['count'] = df['IP_address'].map(df['IP_address'].value_counts())
IP_address count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
|
show 3 more comments
10
10
10
You can use value_counts() with map
df['count'] = df['IP_address'].map(df['IP_address'].value_counts())
IP_address count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
You can use value_counts() with map
df['count'] = df['IP_address'].map(df['IP_address'].value_counts())
IP_address count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
edited Nov 21 '18 at 17:55
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
answered Sep 20 '17 at 20:29
VaishaliVaishali
20.8k41132
20.8k41132
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
|
show 3 more comments
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
1
1
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
I like your solution more, compared to mine... :)
– MaxU
Sep 20 '17 at 20:32
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - Thanks a lot. This has worked.
– Doubt Dhanabalu
Sep 20 '17 at 20:33
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
@Vaishali - I have one question. The resultant value is a float. Should i make something here to convert to integer or should i take that as a separate code?
– Doubt Dhanabalu
Sep 20 '17 at 20:39
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
Shouldn't be. When I try df.dtypes, I get IP_address object, count int64
– Vaishali
Sep 20 '17 at 20:40
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
oh, ok, I got to be float64.
– Doubt Dhanabalu
Sep 20 '17 at 20:42
|
show 3 more comments
9
Using pd.factorize
This should be a very fast solution that scales well for large data
f, u = pd.factorize(df.IP_address.values)
df.assign(IP_address_Count=np.bincount(f)[f])
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
add a comment |
9
Using pd.factorize
This should be a very fast solution that scales well for large data
f, u = pd.factorize(df.IP_address.values)
df.assign(IP_address_Count=np.bincount(f)[f])
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
add a comment |
9
9
9
Using pd.factorize
This should be a very fast solution that scales well for large data
f, u = pd.factorize(df.IP_address.values)
df.assign(IP_address_Count=np.bincount(f)[f])
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
Using pd.factorize
This should be a very fast solution that scales well for large data
f, u = pd.factorize(df.IP_address.values)
df.assign(IP_address_Count=np.bincount(f)[f])
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
answered Sep 20 '17 at 20:48
piRSquaredpiRSquared
156k22150294
156k22150294
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
add a comment |
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
1
1
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
Yes, it is quick . .. currently , I am using this method for count unique ;-)
– Wen-Ben
Sep 20 '17 at 20:50
add a comment |
8
NumPy way -
tags, C = np.unique(df.IP_address, return_counts=1, return_inverse=1)[1:]
df['IP_address_Count'] = C[tags]
Sample output -
In [275]: df
Out[275]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
add a comment |
8
NumPy way -
tags, C = np.unique(df.IP_address, return_counts=1, return_inverse=1)[1:]
df['IP_address_Count'] = C[tags]
Sample output -
In [275]: df
Out[275]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
add a comment |
8
8
8
NumPy way -
tags, C = np.unique(df.IP_address, return_counts=1, return_inverse=1)[1:]
df['IP_address_Count'] = C[tags]
Sample output -
In [275]: df
Out[275]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
NumPy way -
tags, C = np.unique(df.IP_address, return_counts=1, return_inverse=1)[1:]
df['IP_address_Count'] = C[tags]
Sample output -
In [275]: df
Out[275]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
answered Sep 20 '17 at 20:28
DivakarDivakar
156k1487178
156k1487178
add a comment |
add a comment |
7
In [75]: df['IP_address_Count'] = df.groupby('IP_address')['IP_address'].transform('size')
In [76]: df
Out[76]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
add a comment |
7
In [75]: df['IP_address_Count'] = df.groupby('IP_address')['IP_address'].transform('size')
In [76]: df
Out[76]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
add a comment |
7
7
7
In [75]: df['IP_address_Count'] = df.groupby('IP_address')['IP_address'].transform('size')
In [76]: df
Out[76]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
In [75]: df['IP_address_Count'] = df.groupby('IP_address')['IP_address'].transform('size')
In [76]: df
Out[76]:
IP_address IP_address_Count
0 IP1 3
1 IP1 3
2 IP1 3
3 IP4 5
4 IP4 5
5 IP4 5
6 IP4 5
7 IP4 5
8 IP7 3
9 IP7 3
10 IP7 3
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
answered Sep 20 '17 at 20:28
MaxUMaxU
122k12121173
122k12121173
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
add a comment |
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
thank you max for taking time and answering.
– Doubt Dhanabalu
Sep 20 '17 at 20:34
add a comment |
1
ip_set = df.IP_address.unique()
dict_temp = {}
for ip in ip_set:
dict_temp[ip] = df[df.IP_address == ip].IP_address.value_counts()[0]
df['counts'] = [dict_temp[ip] for ip in df.IP_address]
This seems to give me the sort of output that you desire
EDIT: Vaishali's use of map is perfect
answered Sep 20 '17 at 20:41
NRKNRK
363
add a comment |
1
ip_set = df.IP_address.unique()
dict_temp = {}
for ip in ip_set:
dict_temp[ip] = df[df.IP_address == ip].IP_address.value_counts()[0]
df['counts'] = [dict_temp[ip] for ip in df.IP_address]
This seems to give me the sort of output that you desire
EDIT: Vaishali's use of map is perfect
answered Sep 20 '17 at 20:41
NRKNRK
363
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ip_set = df.IP_address.unique()
dict_temp = {}
for ip in ip_set:
dict_temp[ip] = df[df.IP_address == ip].IP_address.value_counts()[0]
df['counts'] = [dict_temp[ip] for ip in df.IP_address]
This seems to give me the sort of output that you desire
EDIT: Vaishali's use of map is perfect
answered Sep 20 '17 at 20:41
NRKNRK
363
ip_set = df.IP_address.unique()
dict_temp = {}
for ip in ip_set:
dict_temp[ip] = df[df.IP_address == ip].IP_address.value_counts()[0]
df['counts'] = [dict_temp[ip] for ip in df.IP_address]
This seems to give me the sort of output that you desire
EDIT: Vaishali's use of map is perfect
answered Sep 20 '17 at 20:41
NRKNRK
363
answered Sep 20 '17 at 20:41
NRKNRK
363
answered Sep 20 '17 at 20:41
NRKNRK
363
answered Sep 20 '17 at 20:41
NRKNRK
363
363
add a comment |
add a comment |
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