I load CSV document with different number of columns. Therefore I got this error:
Expected 12 fields in line 29, saw 13
To avoid this error I use the hack names=range(24)
df = pd.read_csv(filename, header=None, quoting=csv.QUOTE_NONE, dtype='object', sep=data_file_delimiter, engine='python', encoding = "utf-8", names=range(24))
Problem is I need to know the real number of columns to group this data further into dict data:
data = {}
for row in df.rows:
line = line.strip()
row = line.split(' ')
if len(row) not in data:
data[ len(row) ] = []
data[ len(row) ].append(row)
You can have the number of columns using len(df.columns) but if you only want to convert a pandas df to a dictionary then there are already many built-in methods as given below,
df = pd.DataFrame({'col1': [1, 2], 'col2': [0.5, 0.75]},index=['row1', 'row2'])
df
col1 col2
row1 1 0.50
row2 2 0.75
df.to_dict()
{'col1': {'row1': 1, 'row2': 2}, 'col2': {'row1': 0.5, 'row2': 0.75}}
# You can specify the return orientation.
df.to_dict('series')
{'col1': row1 1
row2 2
Name: col1, dtype: int64,
'col2': row1 0.50
row2 0.75
Name: col2, dtype: float64}
df.to_dict('split')
{'index': ['row1', 'row2'], 'columns': ['col1', 'col2'],
'data': [[1, 0.5], [2, 0.75]]}
df.to_dict('records')
[{'col1': 1, 'col2': 0.5}, {'col1': 2, 'col2': 0.75}]
df.to_dict('index')
{'row1': {'col1': 1, 'col2': 0.5}, 'row2': {'col1': 2, 'col2': 0.75}}
df.to_dict('tight')
{'index': ['row1', 'row2'], 'columns': ['col1', 'col2'],
'data': [[1, 0.5], [2, 0.75]], 'index_names': [None], 'column_names': [None]}
# You can also specify the mapping type.
from collections import OrderedDict, defaultdict
df.to_dict(into=OrderedDict)
OrderedDict([('col1', OrderedDict([('row1', 1), ('row2', 2)])),
('col2', OrderedDict([('row1', 0.5), ('row2', 0.75)]))])
Taken from here
I have a data frame with number of columns I want to group them under two main groups A and B while preserving the old columns names as dictionary as follow
index userid col1 col2 col3 col4 col5 col6 col7
0 1 6 3 Nora 100 11 22 44
the desired data frame is as follow
index userid A B
0 1 {"col1":6, "col2":3, "col3":"Nora","col4":100} {"col5":11, "col6":22, "col7":44}
To match your desired dataframe exactly:
>>> import pandas as pd
# recreating your data
>>> df = pd.DataFrame.from_dict({'index': [0], 'userid': [1], 'col1': [6], 'col2': [3], 'col3': ['Nora'], 'col4': [100], 'col5': [11], 'col6': [22], 'col7': [44]})
# copy of unchanged columns
>>> df_new = df[['index', 'userid']].copy()
# grouping columns together
>>> df_new['A'] = df[['col1', 'col2', 'col3', 'col4']].copy().to_dict(orient='records')
>>> df_new['B'] = df[['col5', 'col6', 'col7']].copy().to_dict(orient='records')
>>> df_new
index userid A B
0 0 1 {'col1': 6, 'col2': 3, 'col3': 'Nora', 'col4': 100} {'col5': 11, 'col6': 22, 'col7': 44}
You can try something like this:
d = {'col1': 'A',
'col2': 'A',
'col3': 'A',
'col4': 'A',
'col5': 'B',
'col6': 'B',
'col7': 'B'}
df.groupby(d, axis=1).apply(pd.DataFrame.to_dict, orient='series').to_frame().T
Output:
A B
0 {'col1': [6], 'col2': [3], 'col3': ['Nora'], '... {'col5': [11], 'col6': [22], 'col7': [44]}
Working with the original dataframe.
import pandas as pd
df1 = pd.DataFrame({'index':[0], 'userid':[1],
'col1': [6], 'col2': [3], 'col3': ['Nora'] ,'col4':[100],
'col5':[11], 'col6': [22], 'col7':[44]})
df1['A'] = df1[['col1', 'col2', 'col3', 'col4']].to_dict(orient='records')
df1['B'] = df1[['col5', 'col6', 'col7']].to_dict(orient='records')
df1.drop(df1.columns[range(2, 9)], axis=1, inplace=True)
print(df1)
I'm currently trying to do something very similar to:
for letter in ['a', 'b', 'c']:
key1 = f'{letter}_1'
key2 = f'{letter}_2'
numbers = {
key1: 1,
key2: 2
}
I would expect numbers to be: {'a_1': 1, 'a_2': 2, 'b_1': 1, 'b_2': 2, 'c_1': 1, 'c_2': 2}. Instead I get: {'c_1': 1, 'c_2': 2}.
How can I go about producing the former?
I think the issue is that you did not initialise the dict before the for loop.
numbers = {}
for letter in ['a', 'b', 'c']:
key1 = f'{letter}_1'
key2 = f'{letter}_2'
numbers.update ({
key1: 1,
key2: 2
})
print(numbers)
You could try doing something like this
numbers = {}
for letter in ['a', 'b', 'c']:
key1 = f'{letter}_1'
key2 = f'{letter}_2'
numbers.update({
key1: 1,
key2: 2
})
You need to initialize your dictionary outside the for-loop. In your code, you're creating a new dictionary with each iteration.
You are creating a new object every loop.
numbers = {}
for letter in ['a', 'b', 'c']:
key1 = f'{letter}_1'
key2 = f'{letter}_2'
numbers.update({
key1: 1,
key2: 2
})
You could, if desired, build it with a dict comprehension as:
mydict = {f'{x}_{str(y)}':y for x in ['a','b','c'] for y in (1,2)}
Which gives:
{'a_1': 1, 'a_2': 2, 'b_1': 1, 'b_2': 2, 'c_1': 1, 'c_2': 2}
It's a bit hard to read but not too bad.
I'm trying to change the values of only certain values in a dataframe:
test = pd.DataFrame({'col1': ['a', 'a', 'b', 'c'], 'col2': [1, 2, 3, 4]})
dict_curr = {'a':2}
test['col2'] = np.where(test.col1 == 'a', test.col1.map(lambda x: dict_curr[x]), test.col2)
However, this doesn't seem to work because even though I'm looking only at the values in col1 that are 'a', the error says
KeyError: 'b'
Implying that it also looks at the values of col1 with values 'b'. Why is this? And how do I fix it?
The error is originating from the test.col1.map(lambda x: dict_curr[x]) part. You look up the values from col1 in dict_curr, which only has an entry for 'a', not for 'b'.
You can also just index the dataframe:
test.loc[test.col1 == 'a', 'col2'] = 2
The problem is that when you call np.where all of its parameters are evaluated first, and then the result is decided depending on the condition. So the dictionary is queried also for 'b' and 'c', even if those values will be discarded later. Probably the easiest fix is:
import pandas as pd
import numpy as np
test = pd.DataFrame({'col1': ['a', 'a', 'b', 'c'], 'col2': [1, 2, 3, 4]})
dict_curr = {'a': 2}
test['col2'] = np.where(test.col1 == 'a', test.col1.map(lambda x: dict_curr.get(x, 0)), test.col2)
This will give the value 0 for keys not in the dictionary, but since it will be discarded later it does not matter which value you use.
Another easy way of getting the same result is:
import pandas as pd
test = pd.DataFrame({'col1': ['a', 'a', 'b', 'c'], 'col2': [1, 2, 3, 4]})
dict_curr = {'a': 2}
test['col2'] = test.apply(lambda x: dict_curr.get(x.col1, x.col2), 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