Pandas create new column based on first unique values of existing column
2
I'm trying to add a new column to a dataframe with only unique values from an existing column. There will be fewer rows in the new column maybe with np.nan values where duplicates would have been.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3,4,5], 'b':[3,4,3,4,5]})
df
a b
0 1 3
1 2 4
2 3 3
3 4 4
4 5 5
Goal:
a b c
0 1 3 3
1 2 4 4
2 3 3 nan
3 4 4 nan
4 5 5 5
I've tried:
df['c'] = np.where(df['b'].unique(), df['b'], np.nan)
It throws: operands could not be broadcast together with shapes (3,) (5,) ()
python python-3.x pandas numpy unique
edited Nov 14 '18 at 17:43
jpp
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
add a comment |
2
I'm trying to add a new column to a dataframe with only unique values from an existing column. There will be fewer rows in the new column maybe with np.nan values where duplicates would have been.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3,4,5], 'b':[3,4,3,4,5]})
df
a b
0 1 3
1 2 4
2 3 3
3 4 4
4 5 5
Goal:
a b c
0 1 3 3
1 2 4 4
2 3 3 nan
3 4 4 nan
4 5 5 5
I've tried:
df['c'] = np.where(df['b'].unique(), df['b'], np.nan)
It throws: operands could not be broadcast together with shapes (3,) (5,) ()
python python-3.x pandas numpy unique
edited Nov 14 '18 at 17:43
jpp
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
add a comment |
2
2
2
2
I'm trying to add a new column to a dataframe with only unique values from an existing column. There will be fewer rows in the new column maybe with np.nan values where duplicates would have been.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3,4,5], 'b':[3,4,3,4,5]})
df
a b
0 1 3
1 2 4
2 3 3
3 4 4
4 5 5
Goal:
a b c
0 1 3 3
1 2 4 4
2 3 3 nan
3 4 4 nan
4 5 5 5
I've tried:
df['c'] = np.where(df['b'].unique(), df['b'], np.nan)
It throws: operands could not be broadcast together with shapes (3,) (5,) ()
python python-3.x pandas numpy unique
edited Nov 14 '18 at 17:43
jpp
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
I'm trying to add a new column to a dataframe with only unique values from an existing column. There will be fewer rows in the new column maybe with np.nan values where duplicates would have been.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3,4,5], 'b':[3,4,3,4,5]})
df
a b
0 1 3
1 2 4
2 3 3
3 4 4
4 5 5
Goal:
a b c
0 1 3 3
1 2 4 4
2 3 3 nan
3 4 4 nan
4 5 5 5
I've tried:
df['c'] = np.where(df['b'].unique(), df['b'], np.nan)
It throws: operands could not be broadcast together with shapes (3,) (5,) ()
python python-3.x pandas numpy unique
python python-3.x pandas numpy unique
edited Nov 14 '18 at 17:43
jpp
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
edited Nov 14 '18 at 17:43
jpp
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
edited Nov 14 '18 at 17:43
jpp
101k2162111
edited Nov 14 '18 at 17:43
jpp
101k2162111
edited Nov 14 '18 at 17:43
jpp
101k2162111
101k2162111
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
asked Nov 14 '18 at 17:36
Derek_PDerek_P
328215
328215
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
3
mask + duplicated
You can use Pandas methods for masking a series:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
add a comment |
2
Use duplicated with np.where:
df['c'] = np.where(df['b'].duplicated(),np.nan,df['b'])
Or:
df['c'] = df['b'].where(~df['b'].duplicated(),np.nan)
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
add a comment |
0
ppg wrote:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
I like the code, but the last column should also give NaN
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 NaN
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
mask + duplicated
You can use Pandas methods for masking a series:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
add a comment |
3
mask + duplicated
You can use Pandas methods for masking a series:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
add a comment |
3
3
3
mask + duplicated
You can use Pandas methods for masking a series:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
mask + duplicated
You can use Pandas methods for masking a series:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
answered Nov 14 '18 at 17:42
jppjpp
101k2162111
101k2162111
add a comment |
add a comment |
2
Use duplicated with np.where:
df['c'] = np.where(df['b'].duplicated(),np.nan,df['b'])
Or:
df['c'] = df['b'].where(~df['b'].duplicated(),np.nan)
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
add a comment |
2
Use duplicated with np.where:
df['c'] = np.where(df['b'].duplicated(),np.nan,df['b'])
Or:
df['c'] = df['b'].where(~df['b'].duplicated(),np.nan)
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
add a comment |
2
2
2
Use duplicated with np.where:
df['c'] = np.where(df['b'].duplicated(),np.nan,df['b'])
Or:
df['c'] = df['b'].where(~df['b'].duplicated(),np.nan)
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
Use duplicated with np.where:
df['c'] = np.where(df['b'].duplicated(),np.nan,df['b'])
Or:
df['c'] = df['b'].where(~df['b'].duplicated(),np.nan)
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
edited Nov 14 '18 at 17:48
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
answered Nov 14 '18 at 17:43
Sandeep KadapaSandeep Kadapa
7,098830
7,098830
add a comment |
add a comment |
0
ppg wrote:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
I like the code, but the last column should also give NaN
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 NaN
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
add a comment |
0
ppg wrote:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
I like the code, but the last column should also give NaN
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 NaN
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
add a comment |
0
0
0
ppg wrote:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
I like the code, but the last column should also give NaN
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 NaN
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
ppg wrote:
df['c'] = df['b'].mask(df['b'].duplicated())
print(df)
a b c
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 5.0
I like the code, but the last column should also give NaN
0 1 3 3.0
1 2 4 4.0
2 3 3 NaN
3 4 4 NaN
4 5 5 NaN
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
answered Nov 14 '18 at 18:03
Michael G.Michael G.
2231316
2231316
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
add a comment |
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
I don't understand your answer / point. Can you explain further?
– jpp
Jan 13 at 14:12
add a comment |
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