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Value assignment between Series using .loc, same size with different indices

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1

I have an issue where I have two series of equal size but with different indices. I want to set one series equal to the other. This can be done by resetting the index, but in this case, I am using .loc to slice a DataFrame to a Series.

Setup

Let's say I have a data frame:

df = pd.DataFrame('name': ['a', 'b', 'a', 'b'], 
 'num': [1, 1, 2, 2], 
 'val': None)

 name num val
0 a 1 None
1 b 1 None
2 a 2 None
3 b 2 None

And two Series with values:

vals_a = pd.Series([21,32])
vals_b = pd.Series([43,54])

Desired Outcome

 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

My Code

df.loc[df.name == 'a', 'vals'] = vals_a
df.loc[df.name == 'b', 'vals'] = vals_b

 name num val vals
0 a 1 None 21.0
1 b 1 None 54.0
2 a 2 None NaN
3 b 2 None NaN

This is a simplified version of my problem. In reality, vals_a and vals_b are slices of a separate DataFrame with variable indices. How do I slice this so my original DataFrame stores the values in the smaller Series?

python pandas

share|improve this question

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

add a comment | 

1

I have an issue where I have two series of equal size but with different indices. I want to set one series equal to the other. This can be done by resetting the index, but in this case, I am using .loc to slice a DataFrame to a Series.

Setup

Let's say I have a data frame:

df = pd.DataFrame('name': ['a', 'b', 'a', 'b'], 
 'num': [1, 1, 2, 2], 
 'val': None)

 name num val
0 a 1 None
1 b 1 None
2 a 2 None
3 b 2 None

And two Series with values:

vals_a = pd.Series([21,32])
vals_b = pd.Series([43,54])

Desired Outcome

 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

My Code

df.loc[df.name == 'a', 'vals'] = vals_a
df.loc[df.name == 'b', 'vals'] = vals_b

 name num val vals
0 a 1 None 21.0
1 b 1 None 54.0
2 a 2 None NaN
3 b 2 None NaN

This is a simplified version of my problem. In reality, vals_a and vals_b are slices of a separate DataFrame with variable indices. How do I slice this so my original DataFrame stores the values in the smaller Series?

python pandas

share|improve this question

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

add a comment | 

I have an issue where I have two series of equal size but with different indices. I want to set one series equal to the other. This can be done by resetting the index, but in this case, I am using .loc to slice a DataFrame to a Series.

Setup

Let's say I have a data frame:

df = pd.DataFrame('name': ['a', 'b', 'a', 'b'], 
 'num': [1, 1, 2, 2], 
 'val': None)

 name num val
0 a 1 None
1 b 1 None
2 a 2 None
3 b 2 None

And two Series with values:

vals_a = pd.Series([21,32])
vals_b = pd.Series([43,54])

Desired Outcome

 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

My Code

df.loc[df.name == 'a', 'vals'] = vals_a
df.loc[df.name == 'b', 'vals'] = vals_b

 name num val vals
0 a 1 None 21.0
1 b 1 None 54.0
2 a 2 None NaN
3 b 2 None NaN

This is a simplified version of my problem. In reality, vals_a and vals_b are slices of a separate DataFrame with variable indices. How do I slice this so my original DataFrame stores the values in the smaller Series?

python pandas

share|improve this question

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

df = pd.DataFrame('name': ['a', 'b', 'a', 'b'], 
 'num': [1, 1, 2, 2], 
 'val': None)

 name num val
0 a 1 None
1 b 1 None
2 a 2 None
3 b 2 None

vals_a = pd.Series([21,32])
vals_b = pd.Series([43,54])

 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

df.loc[df.name == 'a', 'vals'] = vals_a
df.loc[df.name == 'b', 'vals'] = vals_b

 name num val vals
0 a 1 None 21.0
1 b 1 None 54.0
2 a 2 None NaN
3 b 2 None NaN

share|improve this question

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

share|improve this question

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

asked Mar 8 at 21:57

Cilantro DitrekCilantro Ditrek

1941318

2 Answers
2

active

oldest

votes

4

pandas is index sensitive , which means when you assign , beside the condition you mentioned within the loc , it will still check the index match or not

df.loc[df.name == 'a', 'vals'] = vals_a.values
df.loc[df.name == 'b', 'vals'] = vals_b.values
df
Out[964]: 
 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

share|improve this answer

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

add a comment | 

1

The easiest method I can think of is to use map with a dict of iterators:

iter_a, iter_b = (iter(v) for v in (vals_a, vals_b))

mapping = 'a': iter_a, 'b': iter_b
df['name'].map(lambda x: next(mapping[x]))

0 21
1 43
2 32
3 54
Name: name, dtype: int64

share|improve this answer

answered Mar 8 at 22:05

cs95cs95

143k25164249

add a comment | 

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4

pandas is index sensitive , which means when you assign , beside the condition you mentioned within the loc , it will still check the index match or not

df.loc[df.name == 'a', 'vals'] = vals_a.values
df.loc[df.name == 'b', 'vals'] = vals_b.values
df
Out[964]: 
 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

share|improve this answer

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

add a comment | 

4

pandas is index sensitive , which means when you assign , beside the condition you mentioned within the loc , it will still check the index match or not

df.loc[df.name == 'a', 'vals'] = vals_a.values
df.loc[df.name == 'b', 'vals'] = vals_b.values
df
Out[964]: 
 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

share|improve this answer

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

add a comment | 

pandas is index sensitive , which means when you assign , beside the condition you mentioned within the loc , it will still check the index match or not

df.loc[df.name == 'a', 'vals'] = vals_a.values
df.loc[df.name == 'b', 'vals'] = vals_b.values
df
Out[964]: 
 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

share|improve this answer

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

df.loc[df.name == 'a', 'vals'] = vals_a.values
df.loc[df.name == 'b', 'vals'] = vals_b.values
df
Out[964]: 
 name num val vals
0 a 1 None 21.0
1 b 1 None 43.0
2 a 2 None 32.0
3 b 2 None 54.0

share|improve this answer

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

answered Mar 8 at 22:06

Wen-BenWen-Ben

127k83872

1

The easiest method I can think of is to use map with a dict of iterators:

iter_a, iter_b = (iter(v) for v in (vals_a, vals_b))

mapping = 'a': iter_a, 'b': iter_b
df['name'].map(lambda x: next(mapping[x]))

0 21
1 43
2 32
3 54
Name: name, dtype: int64

share|improve this answer

answered Mar 8 at 22:05

cs95cs95

143k25164249

add a comment | 

1

The easiest method I can think of is to use map with a dict of iterators:

iter_a, iter_b = (iter(v) for v in (vals_a, vals_b))

mapping = 'a': iter_a, 'b': iter_b
df['name'].map(lambda x: next(mapping[x]))

0 21
1 43
2 32
3 54
Name: name, dtype: int64

share|improve this answer

answered Mar 8 at 22:05

cs95cs95

143k25164249

add a comment | 

The easiest method I can think of is to use map with a dict of iterators:

iter_a, iter_b = (iter(v) for v in (vals_a, vals_b))

mapping = 'a': iter_a, 'b': iter_b
df['name'].map(lambda x: next(mapping[x]))

0 21
1 43
2 32
3 54
Name: name, dtype: int64

share|improve this answer

answered Mar 8 at 22:05

cs95cs95

143k25164249

iter_a, iter_b = (iter(v) for v in (vals_a, vals_b))

mapping = 'a': iter_a, 'b': iter_b
df['name'].map(lambda x: next(mapping[x]))

0 21
1 43
2 32
3 54
Name: name, dtype: int64

share|improve this answer

answered Mar 8 at 22:05

cs95cs95

143k25164249

answered Mar 8 at 22:05

cs95cs95

143k25164249

answered Mar 8 at 22:05

cs95cs95

143k25164249