Typeerror: Strptime() Argument 1 Must Be Str, Not List

TypeError: strptime() argument 1 must be str, not list

As the other answers suggest, you require a different way, probably map. You may also use pd.to_datetime() and pass it the whole list. And then use the same as x-axis and the wind_speed for y-axis.

import pandas as pd    
timestamp = pd.to_datetime(T[1:])

It will create a DatetimeIndex which you can again format according to your needs, like:

timestamp = timestamp.strftime("%Y%m%d %I:%M:%S")

In one line:

timestamp = pd.to_datetime(T[1:]).strftime("%Y%m%d %I:%M:%S")

After having two lists for timestamp and wind_speed, use may use something like:

import matplotlib.pyplot as plt
fig, ax = plt.subplots(figsize=(13,6))
ax.plot(timestamp, wind_speed)
plt.xticks(rotation=30)
plt.show()

datetime to string with series in pandas

There is no .str accessor for datetimes and you can't do .astype(str) either.

Instead, use .dt.strftime:

>>> series = pd.Series(['20010101', '20010331'])
>>> dates = pd.to_datetime(series, format='%Y%m%d')
>>> dates.dt.strftime('%Y-%m-%d')
0    2001-01-01
1    2001-03-31
dtype: object

See the docs on customizing date string formats here: strftime() and strptime() Behavior.

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For old pandas versions <0.17.0, one can instead can call .apply with the Python standard library's datetime.strftime:

>>> dates.apply(lambda x: x.strftime('%Y-%m-%d'))
0    2001-01-01
1    2001-03-31
dtype: object

strptime() argument 1 must be str, not Series

you are getting startTime in string, so lets first convert that to datetime objects.

import pandas as pd
import requests

historical = requests.get(
    'https://ftx.com/api/markets/usdt-perp/candles?resolution=14400&start_time=1309062800').json()
historical = pd.DataFrame(historical['result'])

historical['startTime'] = pd.to_datetime(
    historical['startTime']).dt.strftime('%d/%m/%y %H')


print(historical)

response : strptime() argument 1 must be str, not datetime.datetime django datetime format problem

strptime takes a string and converts it to a datetime.datetime object. You should not have passed a datetime.datetime to it.

datetime.datetime.strptime("2021-02-19", "%Y-%m-%d")
datetime.datetime(2021, 2, 19, 0, 0)

If you need to convert a datetime.datetime to a string, use strftime:

mytime = datetime.datetime(2022, 2, 19, 14, 23, 51)
mytime.strftime('%Y-%m-%dT%H:%M:%SZ')
'2022-02-19T14:23:51Z'

TypeError : strptime() argument 1 must be str, not None

The unittest has worked correctly in your code - it indicates your try/except is not capturing the correct exception. Change your exception to TypeError and it will be caught and the message returned.

You should use self.assertRaisesMesssage if your aim is to check the expected error and error message that passing in None as arg1 produces.

mock_val = None
parameter_name = 'Foo'
date_format_info = 'YYYY/MM/DD HH:mm:ss'
expected_msg = f"Invalid value {mock_val} of parameter {parameter_name}. {parameter_name.capitalize()} should be at format '{date_format_info}'"

with self.assertRaisesMessage(TypeError, expected_msg):
    validate_date(self.logger, mock_val, parameter_name, '%Y/%m/%d %H:%M:%S', date_format_info)

TypeError: strptime() argument 1 must be str, not Period

If you take the year through the dt acessor of timeseries, you get integers (instead of "Period" objects):

df['timestamp'] = pd.to_datetime(df['timestamp'], unit='s')
df['timestamp'] = df['timestamp'].dt.year
y1 = df['timestamp'].iloc[0]
y2 = df['timestamp'].iloc[1]
# d1 = datetime.strptime(y1, "%Y") <- No need to recast to datetime!
# d2 = datetime.strptime(y2, "%Y")
diff = abs((y2 - y1))
print(diff)
>>> 14

As you see, I commented the two lines were you were trying to recast the years into datetime objects. Was there a reason for this? From your question, I assumed you wanted the difference in number of years. If you wanted the exact number of days between the timestamps then this should do: (no need to cast and recast):

df['timestamp'] = pd.to_datetime(df['timestamp'], unit='s')
y1 = df['timestamp'].iloc[0]
y2 = df['timestamp'].iloc[1]
diff = abs((y2 - y1).days)
print(diff)
>>> 5122

strptime() argument 1 must be str, not Series time series convert

You can do it in two ways:

Method 1:

Here we pass a string to the function using map

list(map(lambda x: datetime.datetime.strptime(x,'%b %d, %Y').strftime('%m/%d/%Y'), old_df['oldDate']))

Method 2:

Here we pass a series

pd.to_datetime(old_df['oldDate'], format='%b %d, %Y')

TypeError: strptime() argument 1 must be str, not datetime.datetime in tweepy

tweet.created_at should be a datetime object already, so if you want to format to string, datetime.strftime(tweet.created_at, '%Y-%m-%dT%H:%M:%SZ')

TypeError: strptime() argument 1 must be str, not Series

You need to apply this conversion operation for every row item of your DataFrame, therefore use df.apply function e.g. to create a new pd.Series object and then assign it to the needed column.

df['inward_date'] = df['inward_date'].apply(lambda x: datetime.strptime(x[:10], "%Y-%m-%d").strftime("%d-%m-%Y"))

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