Suppose I have the dataframe pd.DataFrame({'a':nan, 'b':nan, 'c':{'a':1, 'b':2},{'a':4, 'b':7, 'c':nan}, {'a':nan, 'b':nan, 'c':{'a':6, 'b':7}}). I want to take the values from the keys in the dictionary in column c and parse them into keys a and b.
Expected output is:
a b c
0 1 2 {'a':1, 'b':2}
1 4 7 nan
2 6 7 {'a':6, 'b':7}
I know how to do this to create new columns, but that is not the task I need for this, since a and b have relevant information needing updates from c. I have not been able to find anything relevant to this task.
Any suggestions for an efficient method would be most welcome.
** EDIT **
The real problem is that I have the following dataframe, which I reduced to the above (in several, no doubt, extraneous steps):
a b c
0 nan nan [{'a':1, 'b':2}, {'a':6, 'b':7}]
1 4 7 nan
and I need to have output, in as few steps as possible, as per
a b c
0 1 2 {'a':1, 'b':2}
1 4 7 nan
2 6 7 {'a':6, 'b':7}
Thanks!
This works:
def func(x):
d = eval(x['c'])
x['a'] = d['a']
x['b'] = d['b']
return x
df = df.apply(lambda x : func(x), axis=1)
How about this:
for t in d['c'].keys():
d[t] = d['c'][t]
Here is an example:
>>> d = {'a': '', 'b': '', 'c':{'a':1, 'b':2}}
>>> d
{'a': '', 'b': '', 'c': {'a': 1, 'b': 2}}
>>> d.keys()
dict_keys(['a', 'b', 'c'])
>>> d['c'].keys()
dict_keys(['a', 'b'])
>>> for t in d['c'].keys():
... d[t] = d['c'][t]
...
>>> d
{'a': 1, 'b': 2, 'c': {'a': 1, 'b': 2}}
>>>
We can turn it into a function:
>>> def updateDict(dict, sourceKey):
... for targetKey in dict[sourceKey].keys():
... dict[targetKey] = dict[sourceKey][targetKey]
...
>>> d = {'a': '', 'b': '', 'c':{'a':1, 'b':2}}
>>> updateDict(d, 'c')
{'a': 1, 'b': 2, 'c': {'a': 1, 'b': 2}}
>>> d = {'a': '', 'b': '', 'c':{'a':1, 'b':2, 'z':1000}}
>>> updateDict(d, 'c')
{'a': 1, 'b': 2, 'c': {'a': 1, 'b': 2, 'z': 1000}, 'z': 1000}
>>>
I have a dataframe containing id and list of dicts:
df = pd.DataFrame({
'list_of_dicts': [[{'a': 1, 'b': 2}, {'a': 11, 'b': 22}],
[{'a': 3, 'b': 4}, {'a': 33, 'b': 44}]],
'id': [100, 200]
})
and I want to normalize it like this:
id a b
0 100 1 2
0 100 3 4
1 200 11 22
1 200 33 44
This gets most of the way:
pd.concat([
pd.DataFrame.from_dict(item)
for item in df.list_of_dicts
])
but is missing the id column.
I'm most interested in readability.
How about something like this:
d = {
'list_of_dicts': [[{'a': 1, 'b': 2}, {'a': 11, 'b': 22}],
[{'a': 3, 'b': 4}, {'a': 33, 'b': 44}]],
'id': [100, 200]
}
df = pd.DataFrame([pd.Series(x) for ld in d['list_of_dicts'] for x in ld])
id = [[x]*len(l) for l,x in zip(d['list_of_dicts'],d['id'])]
df['id'] = pd.Series([x for l in id for x in l])
EDIT - Here's a simpler version
t = [[('id', i)]+list(l.items()) for i in d['id'] for ll in d['list_of_dicts'] for l in ll]
df = pd.DataFrame([dict(x) for x in t])
And, if you really want the id column first, you can change dict to OrderedDict from the collections module.
This is what I call an incomprehension
pd.DataFrame(
*list(map(list, zip(
*[(d, i) for i, l in zip(df.id, df.list_of_dicts) for d in l]
)))
).rename_axis('id').reset_index()
id a b
0 100 1 2
1 100 11 22
2 200 3 4
3 200 33 44
How can I modify a list value inside dataframes? I am trying to adjust data received by JSON and the DataFrame is as below:
The dataframe has 'multiple dictionary' in one list.
Dataframe df:
id options
0 0 [{'a':1 ,'b':2, 'c':3, 'd':4},{'a':5 ,'b':6, 'c':7, 'd':8}]
1 1 [{'a':9 ,'b':10, 'c':11, 'd':12},{'a':13 ,'b':14, 'c':15, 'd':16}]
2 2 [{'a':9 ,'b':10, 'c':11, 'd':12},{'a':17 ,'b':18, 'c':19, 'd':20}]
If I want to use only 'a' and 'c' key / values in options how can I modify datafames? The expected result would be
Dataframe df:
id options
0 0 [{'a':1 ,'c':3},{'a':5 ,'c':7}]
1 1 [{'a':9, 'c':11},{'a':13,'c':15}]
2 2 [{'a':9 ,'c':11},{'a':17,c':19}]
I tried filtering but I could not assign the value to the dataframe
for x in totaldf['options']:
for y in x:
y = {a: y[a], 'c': y['c']} ...?
Using nested listed comprehension:
df['options'] = [[{'a': y['a'], 'c': y['b']} for y in x] for x in df['options']]
If you wanted to use a for loop it would be something like:
new_options = []
for x in df['options']:
row = []
for y in x:
row.append({a: y[a], 'c': y['c']})
new_options.append(row)
df['options'] = new_options
# An alternative vectorized solution.
df.options = df.options.apply(lambda x: [{k:v for k,v in e.items() if k in['a','c']} for e in x])
Out[398]:
id options
0 0 [{'a': 1, 'c': 3}, {'a': 5, 'c': 7}]
1 1 [{'a': 9, 'c': 11}, {'a': 13, 'c': 15}]
2 2 [{'a': 9, 'c': 11}, {'a': 17, 'c': 19}]
Given the following, how can I set the NaN/None value of the B row based on the other rows? Should I use apply?
d = [
{'A': 2, 'B': Decimal('628.00'), 'C': 1, 'D': 'blue'},
{'A': 1, 'B': None, 'C': 3, 'D': 'orange'},
{'A': 3, 'B': None, 'C': 1, 'D': 'orange'},
{'A': 2, 'B': Decimal('575.00'), 'C': 2, 'D': 'blue'},
{'A': 4, 'B': None, 'C': 1, 'D': 'blue'},
]
df = pd.DataFrame(d)
# Make sure types are correct
df['B'] = df['B'].astype('float')
df['C'] = df['C'].astype('int')
In : df
Out:
A B C D
0 2 628 1 blue
1 1 NaN 3 orange
2 3 NaN 1 orange
3 2 575 2 blue
4 4 NaN 1 blue
In : df.dtypes
Out:
A int64
B float64
C int64
D object
dtype: object
Here is an example of the "rules" to set B when the value is None:
def make_B(c, d):
"""When B is None, the value of B depends on C and D."""
if d == 'blue':
return Decimal('1400.89') * 1 * c
elif d == 'orange':
return Decimal('2300.57') * 2 * c
raise
Here is the way I solve it:
I define make_B as below:
def make_B(x):
if np.isnan(x['B']):
"""When B is None, the value of B depends on C and D."""
if x['D'] == 'blue':
return Decimal('1400.89') * 1 * x['C']
elif x['D'] == 'orange':
return Decimal('2300.57') * 2 * x['C']
else:
return x['B']
Then I use apply:
df.apply(make_B,axis=1)
I have a DataFrame with four columns. I want to convert this DataFrame to a python dictionary. I want the elements of first column be keys and the elements of other columns in same row be values.
DataFrame:
ID A B C
0 p 1 3 2
1 q 4 3 2
2 r 4 0 9
Output should be like this:
Dictionary:
{'p': [1,3,2], 'q': [4,3,2], 'r': [4,0,9]}
The to_dict() method sets the column names as dictionary keys so you'll need to reshape your DataFrame slightly. Setting the 'ID' column as the index and then transposing the DataFrame is one way to achieve this.
to_dict() also accepts an 'orient' argument which you'll need in order to output a list of values for each column. Otherwise, a dictionary of the form {index: value} will be returned for each column.
These steps can be done with the following line:
>>> df.set_index('ID').T.to_dict('list')
{'p': [1, 3, 2], 'q': [4, 3, 2], 'r': [4, 0, 9]}
In case a different dictionary format is needed, here are examples of the possible orient arguments. Consider the following simple DataFrame:
>>> df = pd.DataFrame({'a': ['red', 'yellow', 'blue'], 'b': [0.5, 0.25, 0.125]})
>>> df
a b
0 red 0.500
1 yellow 0.250
2 blue 0.125
Then the options are as follows.
dict - the default: column names are keys, values are dictionaries of index:data pairs
>>> df.to_dict('dict')
{'a': {0: 'red', 1: 'yellow', 2: 'blue'},
'b': {0: 0.5, 1: 0.25, 2: 0.125}}
list - keys are column names, values are lists of column data
>>> df.to_dict('list')
{'a': ['red', 'yellow', 'blue'],
'b': [0.5, 0.25, 0.125]}
series - like 'list', but values are Series
>>> df.to_dict('series')
{'a': 0 red
1 yellow
2 blue
Name: a, dtype: object,
'b': 0 0.500
1 0.250
2 0.125
Name: b, dtype: float64}
split - splits columns/data/index as keys with values being column names, data values by row and index labels respectively
>>> df.to_dict('split')
{'columns': ['a', 'b'],
'data': [['red', 0.5], ['yellow', 0.25], ['blue', 0.125]],
'index': [0, 1, 2]}
records - each row becomes a dictionary where key is column name and value is the data in the cell
>>> df.to_dict('records')
[{'a': 'red', 'b': 0.5},
{'a': 'yellow', 'b': 0.25},
{'a': 'blue', 'b': 0.125}]
index - like 'records', but a dictionary of dictionaries with keys as index labels (rather than a list)
>>> df.to_dict('index')
{0: {'a': 'red', 'b': 0.5},
1: {'a': 'yellow', 'b': 0.25},
2: {'a': 'blue', 'b': 0.125}}
Should a dictionary like:
{'red': '0.500', 'yellow': '0.250', 'blue': '0.125'}
be required out of a dataframe like:
a b
0 red 0.500
1 yellow 0.250
2 blue 0.125
simplest way would be to do:
dict(df.values)
working snippet below:
import pandas as pd
df = pd.DataFrame({'a': ['red', 'yellow', 'blue'], 'b': [0.5, 0.25, 0.125]})
dict(df.values)
Follow these steps:
Suppose your dataframe is as follows:
>>> df
A B C ID
0 1 3 2 p
1 4 3 2 q
2 4 0 9 r
1. Use set_index to set ID columns as the dataframe index.
df.set_index("ID", drop=True, inplace=True)
2. Use the orient=index parameter to have the index as dictionary keys.
dictionary = df.to_dict(orient="index")
The results will be as follows:
>>> dictionary
{'q': {'A': 4, 'B': 3, 'D': 2}, 'p': {'A': 1, 'B': 3, 'D': 2}, 'r': {'A': 4, 'B': 0, 'D': 9}}
3. If you need to have each sample as a list run the following code. Determine the column order
column_order= ["A", "B", "C"] # Determine your preferred order of columns
d = {} # Initialize the new dictionary as an empty dictionary
for k in dictionary:
d[k] = [dictionary[k][column_name] for column_name in column_order]
Try to use Zip
df = pd.read_csv("file")
d= dict([(i,[a,b,c ]) for i, a,b,c in zip(df.ID, df.A,df.B,df.C)])
print d
Output:
{'p': [1, 3, 2], 'q': [4, 3, 2], 'r': [4, 0, 9]}
If you don't mind the dictionary values being tuples, you can use itertuples:
>>> {x[0]: x[1:] for x in df.itertuples(index=False)}
{'p': (1, 3, 2), 'q': (4, 3, 2), 'r': (4, 0, 9)}
For my use (node names with xy positions) I found #user4179775's answer to the most helpful / intuitive:
import pandas as pd
df = pd.read_csv('glycolysis_nodes_xy.tsv', sep='\t')
df.head()
nodes x y
0 c00033 146 958
1 c00031 601 195
...
xy_dict_list=dict([(i,[a,b]) for i, a,b in zip(df.nodes, df.x,df.y)])
xy_dict_list
{'c00022': [483, 868],
'c00024': [146, 868],
... }
xy_dict_tuples=dict([(i,(a,b)) for i, a,b in zip(df.nodes, df.x,df.y)])
xy_dict_tuples
{'c00022': (483, 868),
'c00024': (146, 868),
... }
Addendum
I later returned to this issue, for other, but related, work. Here is an approach that more closely mirrors the [excellent] accepted answer.
node_df = pd.read_csv('node_prop-glycolysis_tca-from_pg.tsv', sep='\t')
node_df.head()
node kegg_id kegg_cid name wt vis
0 22 22 c00022 pyruvate 1 1
1 24 24 c00024 acetyl-CoA 1 1
...
Convert Pandas dataframe to a [list], {dict}, {dict of {dict}}, ...
Per accepted answer:
node_df.set_index('kegg_cid').T.to_dict('list')
{'c00022': [22, 22, 'pyruvate', 1, 1],
'c00024': [24, 24, 'acetyl-CoA', 1, 1],
... }
node_df.set_index('kegg_cid').T.to_dict('dict')
{'c00022': {'kegg_id': 22, 'name': 'pyruvate', 'node': 22, 'vis': 1, 'wt': 1},
'c00024': {'kegg_id': 24, 'name': 'acetyl-CoA', 'node': 24, 'vis': 1, 'wt': 1},
... }
In my case, I wanted to do the same thing but with selected columns from the Pandas dataframe, so I needed to slice the columns. There are two approaches.
Directly:
(see: Convert pandas to dictionary defining the columns used fo the key values)
node_df.set_index('kegg_cid')[['name', 'wt', 'vis']].T.to_dict('dict')
{'c00022': {'name': 'pyruvate', 'vis': 1, 'wt': 1},
'c00024': {'name': 'acetyl-CoA', 'vis': 1, 'wt': 1},
... }
"Indirectly:" first, slice the desired columns/data from the Pandas dataframe (again, two approaches),
node_df_sliced = node_df[['kegg_cid', 'name', 'wt', 'vis']]
or
node_df_sliced2 = node_df.loc[:, ['kegg_cid', 'name', 'wt', 'vis']]
that can then can be used to create a dictionary of dictionaries
node_df_sliced.set_index('kegg_cid').T.to_dict('dict')
{'c00022': {'name': 'pyruvate', 'vis': 1, 'wt': 1},
'c00024': {'name': 'acetyl-CoA', 'vis': 1, 'wt': 1},
... }
Most of the answers do not deal with the situation where ID can exist multiple times in the dataframe. In case ID can be duplicated in the Dataframe df you want to use a list to store the values (a.k.a a list of lists), grouped by ID:
{k: [g['A'].tolist(), g['B'].tolist(), g['C'].tolist()] for k,g in df.groupby('ID')}
Dictionary comprehension & iterrows() method could also be used to get the desired output.
result = {row.ID: [row.A, row.B, row.C] for (index, row) in df.iterrows()}
df = pd.DataFrame([['p',1,3,2], ['q',4,3,2], ['r',4,0,9]], columns=['ID','A','B','C'])
my_dict = {k:list(v) for k,v in zip(df['ID'], df.drop(columns='ID').values)}
print(my_dict)
with output
{'p': [1, 3, 2], 'q': [4, 3, 2], 'r': [4, 0, 9]}
With this method, columns of dataframe will be the keys and series of dataframe will be the values.`
data_dict = dict()
for col in dataframe.columns:
data_dict[col] = dataframe[col].values.tolist()
DataFrame.to_dict() converts DataFrame to dictionary.
Example
>>> df = pd.DataFrame(
{'col1': [1, 2], 'col2': [0.5, 0.75]}, index=['a', 'b'])
>>> df
col1 col2
a 1 0.1
b 2 0.2
>>> df.to_dict()
{'col1': {'a': 1, 'b': 2}, 'col2': {'a': 0.5, 'b': 0.75}}
See this Documentation for details