- 查看版本
import numpy as np
print(np.__version__)
- 創(chuàng)建一維數(shù)組
arr = np.arange(10)
3.創(chuàng)建一個(gè)布爾數(shù)組
np.full((3, 3), True, dtype=bool)
np.ones((3,3), dtype=bool)
4.從一維數(shù)組中提取滿足指定條件的元素毫缆？
arr = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
arr[arr % 2 == 1]
5.用numpy數(shù)組中的另一個(gè)值替換滿足條件的元素項(xiàng)？
arr[arr % 2 == 1] = -1
6.在不影響原始數(shù)組的情況下替換滿足條件的元素項(xiàng)兵睛？
arr = np.arange(10)
out = np.where(arr % 2 == 1, -1, arr)
7.改變數(shù)組的形狀？
arr = np.arange(10)
arr.reshape(2, -1) # Setting to -1 automatically decides the number of cols
- 垂直疊加兩個(gè)數(shù)組？
a = np.arange(10).reshape(2,-1)
b = np.repeat(1, 10).reshape(2,-1)
# Answers# Method 1:
np.concatenate([a, b], axis=0)
# Method 2:
np.vstack([a, b])
# Method 3:
np.r_[a, b]
- 水平疊加兩個(gè)數(shù)組祖很？
a = np.arange(10).reshape(2,-1)
b = np.repeat(1, 10).reshape(2,-1)
# Answers# Method 1:
np.concatenate([a, b], axis=1)
# Method 2:
np.hstack([a, b])
# Method 3:
np.c_[a, b]
- 在無硬編碼的情況下生成numpy中的自定義序列笛丙？
a = np.array([1,2,3])
np.r_[np.repeat(a, 3), np.tile(a, 3)]
- 獲取兩個(gè)numpy數(shù)組之間的公共項(xiàng)？
a = np.array([1,2,3,2,3,4,3,4,5,6])
b = np.array([7,2,10,2,7,4,9,4,9,8])
np.intersect1d(a,b)
- 從一個(gè)數(shù)組中刪除存在于另一個(gè)數(shù)組中的項(xiàng)假颇？
a = np.array([1,2,3,4,5])
b = np.array([5,6,7,8,9])
# From 'a' remove all of 'b'
np.setdiff1d(a,b)
13.得到兩個(gè)數(shù)組元素匹配的位置胚鸯？
a = np.array([1,2,3,2,3,4,3,4,5,6])
b = np.array([7,2,10,2,7,4,9,4,9,8])
np.where(a == b)
14.從numpy數(shù)組中提取給定范圍內(nèi)的所有數(shù)字？
a = np.arange(15)
# Method 1
index = np.where((a >= 5) & (a <= 10))
a[index]
# Method 2:
index = np.where(np.logical_and(a>=5, a<=10))
a[index]# > (array([6, 9, 10]),)
# Method 3: (thanks loganzk!)
a[(a >= 5) & (a <= 10)]
15.創(chuàng)建一個(gè)python函數(shù)來處理scalars并在numpy數(shù)組上工作笨鸡？
def maxx(x, y):
"""Get the maximum of two items"""
if x >= y:
return x
else:
return y
pair_max = np.vectorize(maxx, otypes=[float])
a = np.array([5, 7, 9, 8, 6, 4, 5])
b = np.array([6, 3, 4, 8, 9, 7, 1])
pair_max(a, b)
16.交換二維numpy數(shù)組中的兩列姜钳？
# Input
arr = np.arange(9).reshape(3,3)
arr
# Solution
arr[:, [1,0,2]]
17.交換二維numpy數(shù)組中的兩行？
arr = np.arange(9).reshape(3,3)
# Solution
arr[[1,0,2], :]
18.反轉(zhuǎn)二維數(shù)組的行形耗？
# Input
arr = np.arange(9).reshape(3,3)
arr[::-1]
array([[6, 7, 8],
[3, 4, 5],
[0, 1, 2]])
19.反轉(zhuǎn)二維數(shù)組的列哥桥？
arr = np.arange(9).reshape(3,3)
# Solution
arr[:, ::-1]
20.創(chuàng)建包含5到10之間隨機(jī)浮動(dòng)的二維數(shù)組？
# Input
arr = np.arange(9).reshape(3,3)
# Solution Method 1:
rand_arr = np.random.randint(low=5, high=10, size=(5,3)) + np.random.random((5,3))
# print(rand_arr)
# Solution Method 2:
rand_arr = np.random.uniform(5,10, size=(5,3))
print(rand_arr)
21.在numpy數(shù)組中只打印小數(shù)點(diǎn)后三位激涤？
# Input
rand_arr = np.random.random((5,3))
# Create the random array
rand_arr = np.random.random([5,3])
# Limit to 3 decimal places
np.set_printoptions(precision=3)
rand_arr[:4]
- 通過e式科學(xué)記數(shù)法（如1e10）來打印一個(gè)numpy數(shù)組拟糕？
# Reset printoptions to default
np.set_printoptions(suppress=False)
# Create the random array
np.random.seed(100)
rand_arr = np.random.random([3,3])/1e3
rand_arr
np.set_printoptions(suppress=True, precision=6) # precision is ptional
rand_arr
23.限制numpy數(shù)組輸出中打印的項(xiàng)目數(shù)？
np.set_printoptions(threshold=6)
a = np.arange(15)
24.打印完整的numpy數(shù)組而不截?cái)?/p>
# Input
np.set_printoptions(threshold=6)
a = np.arange(15)
# Solution
np.set_printoptions(threshold=np.nan)
a # 有報(bào)錯(cuò)
25.導(dǎo)入數(shù)字和文本的數(shù)據(jù)集保持文本在numpy數(shù)組中完好無損倦踢？
# Solution
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
names = ('sepallength', 'sepalwidth', 'petallength', 'petalwidth', 'species')
# Print the first 3 rows
iris[:3]
- 從1維元組數(shù)組中提取特定列送滞？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_1d = np.genfromtxt(url, delimiter=',', dtype=None)print(iris_1d.shape)
# Solution:
species = np.array([row[4] for row in iris_1d])
species[:5]
27.將1維元組數(shù)組轉(zhuǎn)換為2維numpy數(shù)組？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_1d = np.genfromtxt(url, delimiter=',', dtype=None)
# Solution:# Method 1: Convert each row to a list and get the first 4 items
iris_2d = np.array([row.tolist()[:4] for row in iris_1d])
iris_2d[:4]
# Alt Method 2: Import only the first 4 columns from source url
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
iris_2d[:4]
28.計(jì)算numpy數(shù)組的均值辱挥，中位數(shù)犁嗅，標(biāo)準(zhǔn)差？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
sepallength = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0])
# Solution
mu, med, sd = np.mean(sepallength), np.median(sepallength), np.std(sepallength)print(mu, med, sd)
29.規(guī)范化數(shù)組般贼，使數(shù)組的值正好介于0和1之間愧哟？
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
sepallength = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0])
# Solution
Smax, Smin = sepallength.max(), sepallength.min()
S = (sepallength - Smin)/(Smax - Smin)# or
S = (sepallength - Smin)/sepallength.ptp() # Thanks, David jeda!
print(S)
31.找到numpy數(shù)組的百分位數(shù)？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
sepallength = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0])
# Solution
np.percentile(sepallength, q=[5, 95])
32.在數(shù)組中的隨機(jī)位置插入值哼蛆？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='object')
# Method 1
i, j = np.where(iris_2d)
# i, j contain the row numbers and column numbers of 600 elements of iris_x
np.random.seed(100)
iris_2d[np.random.choice((i), 20), np.random.choice((j), 20)] = np.nan
# Method 2
np.random.seed(100)
iris_2d[np.random.randint(150, size=20), np.random.randint(4, size=20)] = np.nan
# Print first 10 rows
print(iris_2d[:10])
33.在numpy數(shù)組中找到缺失值的位置蕊梧？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
iris_2d[np.random.randint(150, size=20), np.random.randint(4, size=20)] = np.nan
# Solution
print("Number of missing values: \n", np.isnan(iris_2d[:, 0]).sum())print("Position of missing values: \n", np.where(np.isnan(iris_2d[:, 0])))
34.根據(jù)兩個(gè)或多個(gè)條件過濾numpy數(shù)組？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
# Solution
condition = (iris_2d[:, 2] > 1.5) & (iris_2d[:, 0] < 5.0)
iris_2d[condition]
35.從numpy數(shù)組中刪除包含缺失值的行腮介？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
iris_2d[np.random.randint(150, size=20), np.random.randint(4, size=20)] = np.nan
# Solution# No direct numpy function for this.# Method 1:
any_nan_in_row = np.array([~np.any(np.isnan(row)) for row in iris_2d])
iris_2d[any_nan_in_row][:5]
# Method 2: (By Rong)
iris_2d[np.sum(np.isnan(iris_2d), axis = 1) == 0][:5]
36.找到numpy數(shù)組的兩列之間的相關(guān)性肥矢？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
# Solution 1
np.corrcoef(iris[:, 0], iris[:, 2])[0, 1]
# Solution 2
from scipy.stats.stats import pearsonr
corr, p_value = pearsonr(iris[:, 0], iris[:, 2])
print(corr)
37.查找給定數(shù)組是否具有任何空值？
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
np.isnan(iris_2d).any()
38.在numpy數(shù)組中用0替換所有缺失值叠洗？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='float', usecols=[0,1,2,3])
iris_2d[np.random.randint(150, size=20), np.random.randint(4, size=20)] = np.nan
# Solution
iris_2d[np.isnan(iris_2d)] = 0
iris_2d[:4]
39.在numpy數(shù)組中查找唯一值的計(jì)數(shù)甘改？
# Import iris keeping the text column intact
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
names = ('sepallength', 'sepalwidth', 'petallength', 'petalwidth', 'species')
# Solution# Extract the species column as an array
species = np.array([row.tolist()[4] for row in iris])
# Get the unique values and the counts
np.unique(species, return_counts=True)
40.將數(shù)字轉(zhuǎn)換為分類（文本）數(shù)組？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
names = ('sepallength', 'sepalwidth', 'petallength', 'petalwidth', 'species')
# Bin petallength
petal_length_bin = np.digitize(iris[:, 2].astype('float'), [0, 3, 5, 10])
# Map it to respective category
label_map = {1: 'small', 2: 'medium', 3: 'large', 4: np.nan}
petal_length_cat = [label_map[x] for x in petal_length_bin]
# View
petal_length_cat[:4]
41.從numpy數(shù)組的現(xiàn)有列創(chuàng)建新列灭抑？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris_2d = np.genfromtxt(url, delimiter=',', dtype='object')
# Solution# Compute volume
sepallength = iris_2d[:, 0].astype('float')
petallength = iris_2d[:, 2].astype('float')
volume = (np.pi * petallength * (sepallength**2))/3
# Introduce new dimension to match iris_2d's
volume = volume[:, np.newaxis]
# Add the new column
out = np.hstack([iris_2d, volume])
# View
out[:4]
42.在numpy中進(jìn)行概率抽樣十艾？
# Import iris keeping the text column intact
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
# Solution# Get the species column
species = iris[:, 4]
# Approach 1: Generate Probablistically
np.random.seed(100)
a = np.array(['Iris-setosa', 'Iris-versicolor', 'Iris-virginica'])
species_out = np.random.choice(a, 150, p=[0.5, 0.25, 0.25])
# Approach 2: Probablistic Sampling (preferred)
np.random.seed(100)
probs = np.r_[np.linspace(0, 0.500, num=50), np.linspace(0.501, .750, num=50), np.linspace(.751, 1.0, num=50)]
index = np.searchsorted(probs, np.random.random(150))
species_out = species[index]
print(np.unique(species_out, return_counts=True))
方法2是首選方法，因?yàn)樗鼊?chuàng)建了一個(gè)索引變量腾节，該變量可用于取樣2維表格數(shù)據(jù)忘嫉。
43.按另一個(gè)數(shù)組分組時(shí)獲取數(shù)組的第二大值荤牍？
# Import iris keeping the text column intact
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
# Solution# Get the species and petal length columns
petal_len_setosa = iris[iris[:, 4] == b'Iris-setosa', [2]].astype('float')
# Get the second last value
np.unique(np.sort(petal_len_setosa))[-2]
44.按列對(duì)2D數(shù)組進(jìn)行排序
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
names = ('sepallength', 'sepalwidth', 'petallength', 'petalwidth', 'species')
# Sort by column position 0: SepalLength
print(iris[iris[:,0].argsort()][:20])
45.在numpy數(shù)組中找到最常見的值？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
# Solution:
vals, counts = np.unique(iris[:, 2], return_counts=True)print(vals[np.argmax(counts)])
46.找到第一次出現(xiàn)的值大于給定值的位置庆冕？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
# Solution: (edit: changed argmax to argwhere. Thanks Rong!)
np.argwhere(iris[:, 3].astype(float) > 1.0)[0]
47.將大于給定值的所有值替換為給定的截止值康吵？
# Input
np.set_printoptions(precision=2)
np.random.seed(100)
a = np.random.uniform(1,50, 20)
# Solution 1: Using np.clip
np.clip(a, a_min=10, a_max=30)
# Solution 2: Using np.where
print(np.where(a < 10, 10, np.where(a > 30, 30, a)))
48.從numpy數(shù)組中獲取最大n值的位置？
# Input
np.random.seed(100)
a = np.random.uniform(1,50, 20)
# Solution:
print(a.argsort())# > [18 7 3 10 15]
# Solution 2:
np.argpartition(-a, 5)[:5]# > [15 10 3 7 18]
# Below methods will get you the values.# Method 1:
a[a.argsort()][-5:]
# Method 2:
np.sort(a)[-5:]
# Method 3:
np.partition(a, kth=-5)[-5:]
# Method 4:
a[np.argpartition(-a, 5)][:5]
- 計(jì)算數(shù)組中所有可能值的行數(shù)访递？
# 輸出包含10列晦嵌，表示從1到10的數(shù)字。這些值是各行中數(shù)字的計(jì)數(shù)拷姿。 例如惭载，cell(0，2)的值為2响巢，這意味著數(shù)字3在第一行中恰好出現(xiàn)了2次棕兼。
np.random.seed(100)
arr = np.random.randint(1,11,size=(6, 10))
# Solution
def counts_of_all_values_rowwise(arr2d):
# Unique values and its counts row wise
num_counts_array = [np.unique(row, return_counts=True) for row in arr2d]
# Counts of all values row wise
return([[int(b[a==i]) if i in a else 0 for i in np.unique(arr2d)] for a, b in num_counts_array])
# Printprint(np.arange(1,11))
counts_of_all_values_rowwise(arr)
# Example 2:
arr = np.array([np.array(list('bill clinton')), np.array(list('narendramodi')), np.array(list('jjayalalitha'))])print(np.unique(arr))
counts_of_all_values_rowwise(arr)
50.將數(shù)組轉(zhuǎn)換為平面一維數(shù)組？
# **給定：**
arr1 = np.arange(3)
arr2 = np.arange(3,7)
arr3 = np.arange(7,10)
array_of_arrays = np.array([arr1, arr2, arr3])print('array_of_arrays: ', array_of_arrays)
# Solution 1
arr_2d = np.array([a for arr in array_of_arrays for a in arr])
# Solution 2:
arr_2d = np.concatenate(array_of_arrays)print(arr_2d)
- 在numpy中為數(shù)組生成單熱編碼抵乓？
# **給定：**
np.random.seed(101)
arr = np.random.randint(1,4, size=6)
arr
# > array([2, 3, 2, 2, 2, 1])
# Solution:def one_hot_encodings(arr):
uniqs = np.unique(arr)
out = np.zeros((arr.shape[0], uniqs.shape[0]))
for i, k in enumerate(arr):
out[i, k-1] = 1
return out
one_hot_encodings(arr)
# Method 2:
(arr[:, None] == np.unique(arr)).view(np.int8)
52.創(chuàng)建按分類變量分組的行號(hào)？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
species = np.genfromtxt(url, delimiter=',', dtype='str', usecols=4)
np.random.seed(100)
species_small = np.sort(np.random.choice(species, size=20))
species_small
print([i for val in np.unique(species_small) for i, grp in enumerate(species_small[species_small==val])])
- 根據(jù)給定的分類變量創(chuàng)建組ID靶衍？
# **給定：**
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
species = np.genfromtxt(url, delimiter=',', dtype='str', usecols=4)
np.random.seed(100)
species_small = np.sort(np.random.choice(species, size=20))
species_small
# Solution:
output = [np.argwhere(np.unique(species_small) == s).tolist()[0][0] for val in np.unique(species_small) for s in species_small[species_small==val]]
# Solution: For Loop version
output = []
uniqs = np.unique(species_small)
for val in uniqs: # uniq values in group
for s in species_small[species_small==val]: # each element in group
groupid = np.argwhere(uniqs == s).tolist()[0][0] # groupid
output.append(groupid)
print(output)
- 使用numpy對(duì)數(shù)組中的項(xiàng)進(jìn)行排名灾炭？
np.random.seed(10)
a = np.random.randint(20, size=10)print('Array: ', a)
# Solutionprint(a.argsort().argsort())print('Array: ', a)
55.使用numpy對(duì)多維數(shù)組中的項(xiàng)進(jìn)行排名？
# **給定：**
np.random.seed(10)
a = np.random.randint(20, size=[2,5])print(a)
# Solutionprint(a.ravel().argsort().argsort().reshape(a.shape))
56.在二維numpy數(shù)組的每一行中找到最大值颅眶？
# Input
np.random.seed(100)
a = np.random.randint(1,10, [5,3])
a
# Solution 1
np.amax(a, axis=1)
# Solution 2
np.apply_along_axis(np.max, arr=a, axis=1)
57.計(jì)算二維numpy數(shù)組每行的最小值蜈出？
# Input
np.random.seed(100)
a = np.random.randint(1,10, [5,3])
a
# Solution
np.apply_along_axis(lambda x: np.min(x)/np.max(x), arr=a, axis=1)
58.在numpy數(shù)組中找到重復(fù)的記錄？
# Input
np.random.seed(100)
a = np.random.randint(0, 5, 10)
## Solution# There is no direct function to do this as of 1.13.3
# Create an all True array
out = np.full(a.shape[0], True)
# Find the index positions of unique elements
unique_positions = np.unique(a, return_index=True)[1]
# Mark those positions as False
out[unique_positions] = False
print(out)
59.找出數(shù)字的分組均值涛酗？
# Input
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data'
iris = np.genfromtxt(url, delimiter=',', dtype='object')
names = ('sepallength', 'sepalwidth', 'petallength', 'petalwidth', 'species')
# Solution# No direct way to implement this. Just a version of a workaround.
numeric_column = iris[:, 1].astype('float') # sepalwidth
grouping_column = iris[:, 4] # species
# List comprehension version[[group_val, numeric_column[grouping_column==group_val].mean()] for group_val in np.unique(grouping_column)]
# For Loop version
output = []for group_val in np.unique(grouping_column):
output.append([group_val, numeric_column[grouping_column==group_val].mean()])
output
60.將PIL圖像轉(zhuǎn)換為numpy數(shù)組铡原？
from io import BytesIO
from PIL import Image
import PIL, requests
# Import image from URL
URL = 'https://img2.baidu.com/it/u=982549611,3731122317&fm=26&fmt=auto&gp=0.jpg'
response = requests.get(URL)
# Read it as Image
I = Image.open(BytesIO(response.content))
# Optionally resize
I = I.resize([150,150])
# Convert to numpy array
arr = np.asarray(I)
# Optionaly Convert it back to an image and show
im = PIL.Image.fromarray(np.uint8(arr))
Image.Image.show(im)
61.刪除numpy數(shù)組中所有缺少的值？
a = np.array([1,2,3,np.nan,5,6,7,np.nan])
a[~np.isnan(a)]
62.計(jì)算兩個(gè)數(shù)組之間的歐氏距離商叹？
# Input
a = np.array([1,2,3,4,5])
b = np.array([4,5,6,7,8])
# Solution
dist = np.linalg.norm(a-b)
dist
63.在一維數(shù)組中找到所有的局部極大值(或峰值)燕刻？
a = np.array([1, 3, 7, 1, 2, 6, 0, 1])
doublediff = np.diff(np.sign(np.diff(a)))
peak_locations = np.where(doublediff == -2)[0] + 1
peak_locations
64.從二維數(shù)組中減去一維數(shù)組，其中一維數(shù)組的每一項(xiàng)從各自的行中減去剖笙？
# Input
a_2d = np.array([[3,3,3],[4,4,4],[5,5,5]])
b_1d = np.array([1,2,3])
# Solution
print(a_2d - b_1d[:,None])
65.查找數(shù)組中項(xiàng)的第n次重復(fù)索引卵洗？
x = np.array([1, 2, 1, 1, 3, 4, 3, 1, 1, 2, 1, 1, 2])
n = 5
# Solution 1: List comprehension[i for i, v in enumerate(x) if v == 1][n-1]
# Solution 2: Numpy version
np.where(x == 1)[0][n-1]
66.將numpy的datetime 64對(duì)象轉(zhuǎn)換為datetime的datetime對(duì)象？
# **給定：** a numpy datetime64 object
dt64 = np.datetime64('2018-02-25 22:10:10')
# Solutionfrom datetime import datetime
dt64.tolist()
# or
dt64.astype(datetime)
67.計(jì)算numpy數(shù)組的移動(dòng)平均值弥咪？
# Solution# Source: https://stackoverflow.com/questions/14313510/how-to-calculate-moving-average-using-numpydef moving_average(a, n=3) :
ret = np.cumsum(a, dtype=float)
ret[n:] = ret[n:] - ret[:-n]
return ret[n - 1:] / n
np.random.seed(100)
Z = np.random.randint(10, size=10)print('array: ', Z)# Method 1
moving_average(Z, n=3).round(2)
# Method 2: # Thanks AlanLRH!# np.ones(3)/3 gives equal weights. Use np.ones(4)/4 for window size 4.
np.convolve(Z, np.ones(3)/3, mode='valid') .
68.在給定起始點(diǎn)过蹂、長度和步驟的情況下創(chuàng)建一個(gè)numpy數(shù)組序列？
length = 10
start = 5
step = 3
def seq(start, length, step):
end = start + (step*length)
return np.arange(start, end, step)
seq(start, length, step)
69.填寫不規(guī)則系列的numpy日期中的缺失日期聚至？
# Input
dates = np.arange(np.datetime64('2018-02-01'), np.datetime64('2018-02-25'), 2)print(dates)
# Solution ---------------
filled_in = np.array([np.arange(date, (date+d)) for date, d in zip(dates, np.diff(dates))]).reshape(-1)
# add the last day
output = np.hstack([filled_in, dates[-1]])
output
# For loop version -------
out = []for date, d in zip(dates, np.diff(dates)):
out.append(np.arange(date, (date+d)))
filled_in = np.array(out).reshape(-1)
# add the last day
output = np.hstack([filled_in, dates[-1]])
output
70.從給定的一維數(shù)組創(chuàng)建步長酷勺？
def gen_strides(a, stride_len=5, window_len=5):
n_strides = ((a.size-window_len)//stride_len) + 1
# return np.array([a[s:(s+window_len)] for s in np.arange(0, a.size, stride_len)[:n_strides]])
return np.array([a[s:(s+window_len)] for s in np.arange(0, n_strides*stride_len, stride_len)])
print(gen_strides(np.arange(15), stride_len=2, window_len=4))
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