how to remove key value pair in dictionary based on condition python

Question

I have a dictionary like

  {
    'timestamp': 1677231711722,
    'instrument_name': 'BTC-30JUN23-8000-P',
    'index_price': 23845.81,
    'direction': 'buy',
    'amount': 27.0},
   {
    'timestamp': 1677231711722,
    'instrument_name': 'BTC-30JUN23-8000-P',
    'index_price': 23845.81,
    'direction': 'buy',
    'amount': 2.4},
   {
    'timestamp': 1677231702010,
    'instrument_name': 'BTC-25FEB23-27000-C',
    'index_price': 23844.34,
       'direction': 'buy',
        'amount': 0.1}

I have timestamp as

1677228369641

which is derived from

old_time= round((datetime.datetime.now(timezone.utc) - timedelta(hours=1)).timestamp()*1000)

I want to compare timestamp in dict and old_time and remove the keys and values of that dict which are lesser than that.

How can we do it without using for loop efficiently.

Asked By: Madan

||

Source

Answer 1

Key-value pairs from the dictionary based on the condition of timestamp being less than old_time

old_time = round((datetime.datetime.now(timezone.utc) - timedelta(hours=1)).timestamp() * 1000)

# given dictionary
data = [    {        'timestamp': 1677231711722,        'instrument_name': 'BTC-30JUN23-8000-P',        'index_price': 23845.81,        'direction': 'buy',        'amount': 27.0    },    {        'timestamp': 1677231711722,        'instrument_name': 'BTC-30JUN23-8000-P',        'index_price': 23845.81,        'direction': 'buy',        'amount': 2.4    },    {        'timestamp': 1677231702010,        'instrument_name': 'BTC-25FEB23-27000-C',        'index_price': 23844.34,        'direction': 'buy',        'amount': 0.1    }]

# dictionary comprehension to filter based on timestamp
filtered_data = [d for d in data if d['timestamp'] >= old_time]

# print filtered data
print(filtered_data)

Answered By: Souvik

how to remove key value pair in dictionary based on condition python

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